Math 70200

A few comments on last week's homework:

Problem #1 was straightforward and people did well on it, but in arguing that Lipschitz continuity implied that \( |F'|\leq M \) a.e., several people wrote that \[ | F'(x) | = \lim_{h\to 0}\frac{| F(x+h) - F(x)|}{h}\leq \lim_{h\to 0} \frac{M|h|}{|h|}=M. \] This argument only proves that if the limit exists, then it's bounded by M. So it's only valid if you've already shown that F is absolutely continuous and then pointed out that hence it is in \( BV([a,b]) \) for every \( a < b \) and hence it is a.e. differentiable on every interval and hence everywhere. (It's also fine just to say that since it's absolutely continuous, it's differentiable a.e.)

On problem #2, one of the key points in the problem is a fact that we use a lot in Chapter 3.5 and which probably ought to be its own lemma: the Lebesgue-Radon-Nikodym decomposition of a measure \( \mu_F \) with respect to Lebesgue measure has the form \( F'\,dm + \lambda \). That is, the portion of \( \mu_F \) that's absolutely continuous with respect to Lebesgue measure has density \( F' \). This fact is an immediate consequence of the differentiation theorem at the end of Chapter 3.4. You do not need to reprove it if you use it on the final or a qualifying exam (or in a paper). It's a standard fact that people will know.

To do problem #2, first you need to explain why you can reduce to the case that all \( F_j \) are in NBV, namely that you can shift them and make them right continuous while only changing their derivatives on sets of measure zero, and then their new sum will still be finite and its derivative will differ from that of F by only a set of measure zero. Now we'll be able to talk about measures \( \mu_F \) and \(\mu_{F_j}\), and \( \mu_F \) is the sum of the \( \mu_{F_j} \). The Lebesgue-Radon-Nikodym decomposition of \( \mu_{F_j} \) is \( F'_j\,dm +\lambda_j\), and the absolutely continuous part of the decomposition of \( \mu_F \) is \( F'\,dm +\lambda\). You can then confirm that \( \sum F'_j\,dm +\sum \lambda_j \) is a valid Lebesgue-Radon-Nikodym decomposition of \( \mu_F \), and by uniqueness of the decomposition we obtain \( F'=\sum F'_j \) a.e.

On problem #5, a lot of people struggled with the details of proving separability of \( L^p(\mathbb{R}^n,m) \), though everyone seemed to get the main idea. I didn't think about the details ahead of time, but in general I approve of your using the results about n-dimensional Lebesgue measure from Chapter 2.6. The key fact from here for this problem is that any measurable set with finite measure in \( \mathbb{R}^n \) can be approximated by a finite union of sets that are products of intervals. (Incidentally, many people on this problem used the word rectangle to mean such a set. But rectangles in a σ-algebra \( \mathcal{M}\otimes\mathcal{N} \) are just defined as sets \( M\times N \) for arbitrary \( M\in\mathcal{M} \) and \( N\in\mathcal{N} \), not necessarily intervals.) Here is a sketch of the proof of this: We start with the fact that \( m(E) \) is equal to the infimum of \( \sum m(E_j) \) over all covers \( \cup E_j \) where \( E_j \) is a rectangle, i.e., has the form \( A_1\times\cdots\times A_n \). Thus we can find a finite union of sets of this form with arbitrarily small symmetric difference with E. So it suffices to approximate each rectangle \( A_1\times\cdots\times A_n \) by a finite union of rectangles with intervals for sides. And you can do this by approximating each set \( A_i\subset\mathbb{R} \) by a finite union of intervals.

From Theorem 2.40c, we need to construct our dense subset of functions in \( L^p \). We know that integrable simple functions are dense in \( L^p \). Now take an arbitrary such function, and show that we can arbitrarily approximate it in \( L^p \) by an integrable simple function with rational cofficients. Now take such a function, and show we can arbitrarily approximate such a function in \( L^p \) by a simple function one where the sets are rectangles with intervals as sides; it suffices to show that for any \( E\subset\RR^n \), you can approximate \( \chi_E \) in \( L^p \) by \( \chi_{E'} \) for such a set \( E' \), which follows nearly immediately from Theorem 2.40c. Now show you can approximate \( \chi_{E'} \) for any such set \( E' \) by \( \chi_{E''} \) where \(E''\) is a rectangle whose sides are intervals with rational endpoints. It is a lot simpler to do this one step at a time as described than all at once, because if you can arbitrarily approximate an object in one set by an object in another, which can be arbitrarily approximated by an object in another set, and so on, then you can arbitrarily approximate anything in the first set by anything in the last.

By the way, in the \( L^{\infty} \) part of the problem, many people gave solutions that at various points referred to two different collections of balls with the same notation, namely balls of points in \( \mathbb{R}^n \) with the standard Euclidean metric, and balls of functions in \( L^{\infty} \) with the \( L^{\infty} \) metric.

We've done our last problem set, but I recommend the following problems from Section 6.2 (and am happy to discuss them with you): Folland exercises 20(a), 21, and 22.

I've added two problems from Chapter 6.1 to the homework. This is the last assignment. I'll give some suggested problems from Chapter 6.2 as well, but they won't be collected.

Some commentary on last week's homework on Problem #3: It is possible to give an elementary solution that doesn't use any of this class's theory. But this is not a good solution. The good solution is to define a measure μ placing mass \( 2^{-n} \) on each point \( q_n \), where \( q_1,q_2,\ldots \) is an enumeration of the rationals. (You don't need to use \( 2^{-n} \), just something summable.) Then define \( F(x) \) to be the right-continuous, increasing function associated with μ, i.e., \( F(x) = \mu((-\infty,x]) \). Then F is continuous at all points except the rationals by Folland exercise 1.28 from Homework #3. Really this exercise could have been assigned back then, but the idea of doing it now is that we should all internalize the connection between positive measures and right-continuous increasing functions (and between signed measures and functions in NBV).

We will have one last homework assignment due Tuesday, May 7th. I've posted the assignment now but will be adding a few problems from Section 6.1 after class on May 2nd.

As we discussed in class today, the final needs to be held on the last day of class (Tuesday, May 14) because of the scheduling of the qualifying exams. I've updated the calendar now.

I've uploaded Homework 10. Sorry for the delay.

Update: All videos now uploaded. See you on Thursday. If you want to meet with me tomorrow to discuss homework, etc., please send me an email to set it up.

Update: I've added one more lecture video. I expect to add one more.

Here are the first two lecture videos in place of Tuesday's class. I'll post the rest by Tuesday.

• Section 3.4 wrap-up (27 minutes)
• Section 3.5 introduction (4 minutes)
• Section 3.5, Theorem 3.23 (27 minutes)
• Section 3.5, total variation of a function (17 minutes)

I've posted Homework 9.

Some comments on Homework 7. First, let me make a general comment that if you want to apply Fubini's theorem to switch the order of integration in \( \int\int f(x,y)\,d\mu(x)\,d\mu(y) \), you first need to justify that \( f \in L^1(\mu\times\nu) \). But you can't do this just by computing the double integral, since it could be that the double integral (in whichever order) is finite even though the integral against \( \mu\times\nu \) is not defined. Example, from Folland Exercise 2.48: Let \( f(m,n) \) be \( -1 \) if \( m = n+1\) and 1 if \(m=n \), and consider the double sums \( \sum_{m=1}^{\infty}\sum_{n=1}^\infty f(m,n) \) and \( \sum_{n=1}^{\infty}\sum_{m=1}^\infty f(m,n) \), which we think of as integrals against counting measure on the positive integers. These double integrals are both finite but aren't equal. And f is not in \( L^1(\mathbb{Z}_+^2) \), which has to be true since Fubini's theorem failed.

So, how do you actually justify that \( f \in L^1(\mu\times\nu) \) so that you can apply Fubini's theorem? Apply Tonelli's theorem to compute \( \int |f|\,d(\mu\times\nu) \) by computing a double integral, which you can always do since Tonelli's theorem has no requirements on f besides nonnegativity (and measurability).

• Problem #1: the idea is that we're computing the area under \( f(x) \) in two different ways. Let \( E\subseteq \mathbb{R}\times[0,\infty) \) be this region, i.e., \( E = \{ (x,t)\colon f(x)>t \} \). Since \( \chi_E\colon \mathbb{R}\times[0,\infty)\to\mathbb{R} \) is positive, Tonelli's theorem lets us integrate \( \chi_E(x,t) \) on \( \mathbb{R}\times[0,\infty) \) as a double integral in either order, one of which gives \(\int f(x)\,dx \) and one of which gives \( \int_0^\infty \varphi(t)\,dt \).
• Problem #2: Let's start by computing \( \int_{\mathbb{R}^n} f(x_1)\cdots f(x_n)\,dm^n(x) \). The function \( x\mapsto f(x_1)\cdots f(x_n) \) is in \( L^1(m^n) \) since by Tonelli's theorem, the integral of its absolute value is equal to \[ \int\cdots\int |f(x_1)\cdots f(x_n)|\,dx_1\,\cdots dx_n = \biggl(\int |f(x)|\,dx\biggr)^n<\infty \] since \( f\in L^1(m) \). Thus Fubini's theorem applies and the same calculation shows that \[ \int_{\mathbb{R}^n} f(x_1)\cdots f(x_n)\,dm^n(x) = \biggl(\int f(x)\,dx\biggr)^n. \] For a permutation \( \sigma\in S_n \), let \( A_\sigma = \{x\in\mathbb{R}^n\colon x_{\sigma(1)}<\cdots < x_{\sigma(n)}\} \). The integral \[ \int_{A_{\sigma}}f(x_1)\cdots f(x_n)\,dm^n(x) \] is the same for any σ, since we can rename \(x_{\sigma(j)} \) as \(y_j\) and rewrite the integral as \( \int_A f(y_1)\cdots f(y_n)\,dm^n(y) \). (We aren't doing any sort of transformation here—we're just assigning different symbols in place of \(x_1,\ldots,x_n\) and taking advantage of symmetry of the integrand.) Let \( B = \mathbb{R}^n\setminus\cup_{\sigma\in S_n} A_\sigma \), all points in \( \mathbb{R}^n \) whose components are not all distinct. We now have \[ \begin{aligned} n!\int_{A}f(x_1)\cdots f(x_n)\,dm^n(x) &= \sum_{\sigma\in S^n}\int_{A_{\sigma}}f(x_1)\cdots f(x_n)\,dm^n(x)\\ &= \int_{\mathbb{R}^n\setminus B}f(x_1)\cdots f(x_n)\,dm^n(x).\end{aligned} \] The only thing left to solving the problem is showing that B has measure zero, since then we could replace the final integral with one over the whole space, which we've already computed to be \(\bigl( \int f(x)\,dx \bigr)^n \). The best way is to use Tonelli's theorem again. Let \( B_{ij}=\{x\in\mathbb{R}^n\colon x_i=x_j\} \), and observe that \(B = \cup_{i\neq j}B_{ij} \), so it's enough to show that \(m(B_{ij}) =0 \), and by symmetry \(m(B_{12})=0 \) is enough. We compute \[\begin{aligned} m(B_{12}) &= \int\cdots\int\int \chi_{B_{12}}(x)\,dx_1\,dx_2\cdots dx_n\\ &\leq \int\cdots\int\int \chi_{\{x_2\}}(x_1)\,dx_1\,dx_2\cdots dx_n\\ &= \int\cdots\int 0\,dx_2\cdots dx_n=0.\end{aligned} \]
• Problem #3: Not too much to say about this one. You need to know the trick, which is integrating the sum from \(x=0 \) to \(x=1 \) and showing that the integral is finite by switching the order of the sum and integral. You need to take absolute values and apply Tonelli's theorem (or Proposition 2.15).

Just mentioning that we'll skip Chapter 3.3, in which complex measures are defined. A complex measure obeys the same axioms as a signed measure, except that it maps sets to the complex numbers rather than the extended reals. A complex measure \( \nu \) can be decomposed as \( \nu_r + i\nu_i \) where \(\nu_r \) and \(\nu_i\), the real and imaginary parts of the measure, are finite signed measures. The only thing that's different as compared to signed measures is that you have to define the total variation \( |\nu| \) differently. I am hoping to avoid them and stick to signed measures in Section 3.5.

Here is Homework 8. It's on the shorter side.

I've posted the remaining problems for Homework 7. I should mention that these exercises involve the positive and negative variations \( \nu^+ \) and \( \nu^- \) and the total variation \( |\nu| \) for a signed measure \( \nu \), which we didn't cover today. We will cover them on Tuesday, but they're not complicated concepts and you can read about them on p. 87 of Folland if you want to get started before then.

I've graded the midterm and will return it tomorrow. Here are solutions to the exam, or rather extremely succinct sketches of solutions. Of course your solutions need to be longer, but I want to get across the main idea for anyone who didn't get it. Of course, there is more than one solution to each problem. And it is not so easy to come up with optimal solutions to these problems on the spot (especially #5, of which there are many clumsier solutions possible). Please feel free to ask me in more detail about any of them.

1. The function \( \chi_E \) pulls back sets only to \(E\), \(E^c \), \( X \), and \(\varnothing \). These four sets are all measurable if and only if E is.
2. (a) Fatou. (b) \( f_n = \chi_{(n,n+1)}\). (c) Bound \( |f_n - f| \) by \(2|f|\in L^1\) and apply the DCT.
3. For any k, \(\mu(F) \leq \mu\bigl( \cup_{n=k}^{\infty} E_n \bigr) \) since \(F\subseteq \cup_{n=k}^{\infty} E_n\). And by subadditivity, \[ \mu\biggl(\bigcup_{n=k}^{\infty}E_n\biggr)\leq \sum_{n=k}^{\infty}\mu(E_n),\] which vanishes as \(k\to\infty \) by the summability of \( \sum_{n=1}^{\infty} \mu(E_n) \). (Yes, it really is possible to do this problem without using continuity from above/below, though you certainly can use them.)
   This result says if sets \( (E_n) \) have finite measure, then the set of points that appear in infinitely many of these sets has measure zero. Or in probability world, it says that if the probabilities of a sequence of events are summable, then the probability of infinitely many of the events occurring is 0. This is called the Borel–Cantelli lemma.
4. True for simple functions by translation invariance of Lebesgue measure. True for all functions by approximating by simple functions and applying the Monotone Convergence Theorem or the definition of the Lebesgue integral.
5. By the \( m^* \)-measurability of \( [-n,n] \), we have \( m^*(E) = m^*(E_n) + m^*(E\cap [-n,n]^c) \). Thus it suffices to show that \( m^*(E\cap [-n,n]^c)\to 0\) as \(n\to\infty\). Choose a cover of E by h-intervals \( (a_j,b_j] \) whose lengths sum to a finite value. Then \( \sum_{j=N}^{\infty}(b_j-a_j) \) is smaller than \(\epsilon\) for large enough N. Now choose n large enough that \( [-n,n] \) contains the first \(N-1 \) of the h-intervals. The remaining h-intervals then must cover \( E\cap[-n,n]^c \), proving that its outer measure is at most \(\epsilon\), thus demonstrating that \( m^*(E\cap [-n,n]^c)\to 0\).

Here are some comments on Homework 6 now that I'm done grading it:

Problem #1(a) about continuity of f being equivalent to \( H(x) = h(x) \) doesn't really involve any ideas and is more of an undergraduate-style real analysis problem of understanding how one statement involving limits/quantification is equal to another. It can be approached via the epsilon-delta definition of continuity or by a sequential characterization of continuity. I would have liked to see more care taken getting correct and succinct proofs here. If you're not good at solving an exercise like this, it's a skill to work on, because at least part of the point of qualifying exams is to make sure that everyone can hammer out a proof like this.

Problem (b) was trickier. The goal is to show that \(H=G\) a.e. and \(h=g\) a.e. Let \(P_k\) be the sequence of partitions from the proof of Theorem 2.28. The idea is to show equality of these functions for all x that are not any of the countably many endpoints in \(P_k\) for any k. Let x be such a point. The hard part is thinking carefully about limits as \(k\to\infty\) and as \(\delta\to 0\). First, to show \(G(x)\leq H(x)\), fix any \(\delta>0\). The interval in the partition \(P_k\) containing x is within \([x-\delta,x+\delta]\) for large enough k. Then \(G_{P_k}(x)\leq H_\delta(x)\) for large enough k. Taking a limit in k gives \(G(x)\leq H_\delta(x)\). Since this holds for all \(\delta>0\), we have \(G(x)\leq H(x)\).

To show that \(G(x)\geq H(x)\), we have to consider the \(k\to\infty\) and \(\delta\to 0\) limits in the other order. Fix any k. The interval of \(P_k\) containing x contains \([x-\delta,x+\delta]\) for sufficiently small \(\delta>0\) (this is where we're using that x is not an endpoint of the intervals in \(P_k\) ). Hence \(G_{P_k}(x)\geq H_{\delta}(x)\) for small enough \(\delta>0\). Taking a limit as \(\delta\to 0\), we have \(G_{P_k}(x)\geq H(x)\). This holds for all k, and now taking the limit as \(k\to\infty\) gives \(G(x)\geq H(x)\).

This proves that \(G(x)=H(x)\) for all points x that are not endpoints in any partition \(P_k\), and the proof for \(g(x)=h(x)\) is the same. It's possible to do the \(G(x)\leq H(x)\) and \(G(x)\geq H(x)\) parts together, but I think it's more confusing and obscures the nice structure where each inequality is obtained by taking limits \(k\to\infty\) and \(\delta\to 0\) but in different orders. One thing I want to highlight in this proof is the importance of thinking carefully about the order you take limits in. Relatedly, oftentimes when a proof is ambiguous or just hard to follow, the problem is insufficient care about quantification. Are you considering an arbitrary \(\delta\) and then choosing k? Or are you considering an arbitrary k and then choosing \(\delta\)? In a nicely written proof, it should be clear to the reader (and to yourself!) which you're doing.

On Problem #2, the fundamental idea is that continuity of F at x is equivalent to \(\int_x^y f(t)\,dt\) being small when y is close to x, and that this can be proven by applying the dominated convergence theorem to show that \(\int_x^y f(t)\,dt\to 0\) as \(y\to x\). The shortest way to write this proof is to use a sequential characterization of continuity, considering an arbitrary sequence \( (x_n )\) converging to x and trying to show that \( F(x_n)\to F(x)\). Then \(F(x_n)=\int f\chi_{(-\infty,x_n)}\,dm\), and \(f\chi_{(-\infty,x_n)}\) converges to \(f\chi_{(-\infty,x)}\) or \(f\chi_{(-\infty,x]}\) as \(n\to\infty\), which are equal a.e. and integrate to \(F(x)\). So we just need to justify that \(\lim_{n\to\infty}\int f\chi_{(-\infty,x_n)}\,dm = \int \lim_{n\to\infty}f\chi_{(-\infty,x_n)}\,dm\), which follows from the DCT since the integrands are all bounded by integrable function \( |f| \).

In Problem #3, the easiest solution is to choose a subsequence \(f_{n_k}\) so that \( \int f_{n_k}\to\liminf_{n\to\infty} f_n\). Then choose a further subsequence on which the functions converge a.e. to f and then apply Fatou's lemma to this subsequence. It's also possible to work directly with the definition of convergence in measure and avoid use of the result about existence of a subsequence converging a.e. The proof like this in the end is not complicated, but it's definitely a challenge to get it this way.

In Problem #4(a), either you can reproduce the proof of the DCT via Fatou's lemma using #3 in place of the traditional Fatou's lemma, or you can use subsequences again, carefully: Take an arbitrary subsequence \( (f_{n_k})\). It still converges in measure to f, so it has a further subsequence \(f_{m_k}\) converging to f a.e. Now each \(f_{m_k}\) is dominated by \(g\in L^1\), so \(\int f_{m_k}\to \int f \) by the DCT. So, what we've shown here is that every subsequence of \( \int f_n\) has a further subsequence converging to \(\int f\). And this implies that the entire sequence \( \int f_n\) converges to \(\int f\). This is a common strategy that you should file away for future use: If you want to show that \(a_n\to a\), it's enough to show that every subsequence of \(a_n\) has a further subsequence converging to a.

On Problem #5, the main idea is to split the domain of the function up into three parts: the set \(E_0 = \{x\colon f(x)=0\}\), the set \(E_1=\{x\colon 0<|f(x)|\leq 1\}\), and the set \(E_2=\{x\colon |f(x)|>1\}\). We then have \(\int_{E_0}|f|^{1/n}\,d\mu=0\) for all n and we have \(\int _{E_1} |f|^{1/n}\,d\mu\to \mu(E_1)\) by the monotone convergence theorem. On \(E_2\), things are trickier. If \(|f|^{1/n}\) restricted to \(E_2\) is in \(L^1\) for any n, then we can apply the DCT using \(|f|^{1/n}\) for the first such n as the dominating function to obtain \(\int_{E_2}|f|^{1/n}\,d\mu\to \mu(E_2)\). But it could happen that \(|f|^{1/n}\) on \(E_2\) is never in \(L^1\) and \(\int_{E_2}|f|^{1/n}\,d\mu=\infty\) for all n even though \(\mu(E_2)<\infty\). For example, consider \(f(x)=e^{1/x}\) on \([0,1]\).

I've posted the beginning of the next homework assignment, which is due in two weeks. I will add more problems to it next week.

Just a quick comment about the schedule with next week's midterm:

• The midterm will cover Sections 1.1–2.4. We'll be covering Section 2.5 (mostly) this week, but it won't be on the midterm since you won't have had the chance to process it yet by doing homework on it.
• The next homework assignment will not be due until March 28. I'll post its first few questions this Thursday, though, in case you want to get started on it.

Update: Okay, problem statements all posted now.

I've posted the next homework assignment, albeit without the problem statements copied over from Folland. I'll put them in later but didn't want to delay posting the homework until I did it.

Commentary on this week's homework:

• Most people solved Problem #2 by expressing the set where \( \lim_{n\to\infty}f_n(x) \) exists as the set where \( \limsup_{n\to\infty} f_n(x) - \liminf_{n\to\infty}f_n(x) =0\) and then using the result from Folland that these two functions are measurable. This is a fine solution, but there's the issue that \(\limsup_{n\to\infty} f_n(x)\) might be equal to \( \infty \) and \(\liminf_{n\to\infty} f_n(x)\) might be equal to \(-\infty\), and it needs to be defined what the difference between these two functions is in this case. This is dealt with in Folland exercise 2.2(b), which I didn't assign, and which basically states that everything is fine. You could either use that fact or deal with this case separately by hand, but something needs to be said.
  I didn't actually mean for this problem to be quite so finicky—I just had a longer (and worse) solution in mind, namely expressing the set as the set of points where the sequence \(f_n(x)\) is Cauchy, which you can write via countable unions and intersections. This solution evades the whole issue, at least if you interpret \( \lim_{n\to\infty}f_n(x) \) as not existing for an infinite limit. Next time I teach this class I will just not assign this problem, I think.
• There were some common features in people's solutions to Problems #5 and #6 that betray some continued discomfort with Lebesgue measurability vs. Borel measurability.
  On #5, the idea (which basically everyone got) is to express a Lebesgue-measurable set E as the union of Borel B and Lebesgue null set N. Then \(f^{-1}(E)\) is Borel by the continuity of f, and you just need to show that \(f^{-1}(N)\) is Lebesgue measurable.
  Before we talk about how to do that, why is it true that E can be decomposed into B and N? Well, one way to see it is that Lebesgue measure is the completion of its restriction to the Borel sets. By definition of completion (see the end of Chapter 1.3), every Lebesgue-measurable set is then the union of a Borel set together with a subset of a null Borel set (which is therefore a null Lebesgue set). But this is only proven in the exercises of Chapter 1.4. For a slightly less general explanation (only slightly! this works for any of our measures on the real numbers induced by an increasing, right-continuous function), observe that if E has finite Lebesgue measure, we can cover it by a Borel set \(B_n\) such that \( m(B_n) < m(E)+\frac{1}{n} \), either using outer regularity or the outer measure construction of Lebesgue measure. Then \(B:=\cap_n B_n\) is Borel, contains E, and has the same measure as it, showing that \(B\cap E\) has measure zero. This assumed that E had finite measure, and now we just need to remove this restriction by applying σ-finiteness.
  So anyway, back to showing that \(f^{-1}(N)\) is Lebesgue measurable. Many people said that N was contained in a null Borel set \(N'\), then covered \(N'\) by a set of h-intervals whose measure summed to less than \(\epsilon\), then used the bi-Lipschitz property to argue that the pullback of this cover had measure at most \( C\epsilon\), and therefore \(f^{-1}(N')\) was a Borel set of measure zero and hence its subset \( f^{-1}(N) \) is a Lebesgue-null set. There's nothing wrong with this argument, but there's no need to invoke \(N'\). You can just say that N is a Lebesgue-null set and can be covered by h-intervals exactly as \(N'\) could be. Since the pullback of this cover is bounded by \(C\epsilon\), the Lebesgue outer measure of \(f^{-1}(N)\) is bounded by \(C\epsilon\) for any \(\epsilon>0\) and is hence equal to 0. Therefore \(f^{-1}(N)\) is Lebesgue measurable with measure zero (all sets of Lebesgue outer measure zero are Lebesgue measurable).
• On Problem #6, here's a proof. We are told that \(f=g\) μ-a.e., which by definition means that \(f=g\) on a μ-measurable set F such that \(\mu(F^c)=0\). For any measurable set A in the codomain, we express \(g^{-1}(A)\) as \[ g^{-1}(A) = \bigl(g^{-1}(A)\cap F\bigr) \cup \bigl(g^{-1}(A)\cap F^c\bigr)=\bigl(f^{-1}(A)\cap F\bigr) \cup \bigl(g^{-1}(A)\cap F^c\bigr) \] Now \(f^{-1}(A)\cap F\) is measurable since f is measurable and hence \(f^{-1}(A)\) is measurable and we know that F is measurable. And \(g^{-1}(A)\cap F^c\) is measurable since it's a subset of the μ-null set \(F^c\), and we have assumed that μ is complete. Note that we do not need to make any assumptions whatsoever about the codomain of f, though I didn't take off any points if you assumed it was a real-valued function or whatever.
  Nearly everyone—I think all but one person—used basically this argument but with the set F equal to the set where f and g disagree. The problem is that we are not told that this set is measurable (though it is, by completeness of μ, since it's a subset of our set Lebesgue null set F). This really just comes down to understanding precisely what is meant when it's stated that \(f=g\) μ-a.e.

I've posted a (short!) homework assignment for next week.

Here's some commentary on Homework 3 now that I'm done grading it:

• When a measure μ is obtained by restricting an outer measure \( \mu^* \) (as is the case for typical measures we work with), problems can often be solved by working with \(\mu^*\) using covers by elementary sets, or they can be solved by working with μ and using continuity from above and below or inner and outer regularity (see Theorem 1.18, and I realize we hadn't covered this yet when you submitted your homework). I wanted to mention that when you have a choice (e.g., in problems 1 and 2 on Homework 3), it's usually simpler to work with μ.
• I noticed a bit of confusion about the various results we have about extensions of measures and outer measures. Here are some correct statements. I'll just give them informally but will refer to the rigorous statement in Folland:
  • If you construct an outer measure from a premeasure using the cover construction, it agrees with the premeasure on the original algebra (Proposition 1.13).
  • Let \(\mu^*\) be the outer measure constructed from a premeasure \(\mu_0\) on an algebra \(\mathcal{A}\). Assuming your space is σ-finite, there is a unique extension of \(\mu_0\) to \(\sigma(A)\), namely \(\mu^*|_{\sigma(\mathcal{A})}\). That is, \(\mu^*\) is a measure when restricted to the σ-algebra generated by \(\mathcal{A}\), and there is no other measure on \(\sigma(\mathcal{A})\) that agrees with \(\mu_0\) on \(\mathcal{A}\). (There is a weaker version of this statement that holds on spaces that aren't σ-finite, but I want to emphasize the σ-finite version since that's what you'll typically care about.)
• Another source of confusion was that in Problem 5, we construct outer measures from \(\mu_1\) and \(\mu_2\), which are measures, not just premeasures. This is the case where we have a premeasure on an algebra that also happens to be a σ-algebra, which means that the premeasure is also a measure. In this case, the first extension result above says that \(\mu_i^*|_{\mathcal{M}_i} = \mu_i\), while the second extension result has no content whatsoever, just saying that any measure that agrees with \(\mu_i\) on its domain \(\mathcal{M}_i=\sigma(\mathcal{M}_i)\) is equal to \(\mu_i\).
• People's solutions to Problem 1 tended to be long and overcomplicated. For (a), we want to show that for Borel set E, we have \(\mu_F(A)\) equal to 0 or 1 depending on whether E contains 0. First prove that \(\mu_F(\{0\})=0\) by noting that \(\mu_F((-\frac{1}{n},0])=1\) and applying continuity from above (or citing Problem 2). Then observe that \(\mu_F(\{0\}^c)=0\) by a similar application of continuity from below. Now if E doesn't contain 0, then \(E\subseteq\{0\}^c\) and by monotonicity \(\mu_F(E)=0\). And if E does contain 0, then \(\{0\}\subseteq E\subseteq X\), and \(\mu_F(\{0\})=1\) and \(\mu_F(X)=\mu_F(\{0\}^c)+\mu_F(\{0\})=1\), and by monotonicity \(\mu_F(\{0\})=1\). Using outer measures instead is slightly more complicated but still not that bad (see the solution to part (b) posted last week, which includes a solution to (a) using outer measure).

Just letting you know I've posted the next homework assignment, due on Thursday, February 29th.

In talking to people about last week's homework, I've perceived a lot of confusion about Problem 1(b). I haven't graded the homework yet so I'm not sure if your solutions had the right idea or not, but I wanted to say something about it now.

In this question, we have the measure \( \mu_F \) where F takes value 0 on \( [-\infty,0) \) and value 1 on \( [1,\infty) \). I think everybody figured out that this measure is a point mass at 0, i.e., it assigns measure 1 to sets that include 0 and measure 0 to sets that don't include 0.

The question was to figure out which sets are \( \mu^* \)-measurable, where \( \mu^* \) is the outer measure used to construct \( \mu_F \). The answer is that all subsets of \(\mathbb{R}\) are \(\mu^*\)-measurable. I can think of two ways to approach this question. First, we could directly use the definition of \( \mu^* \)-measurable, that a set \( A\subseteq \mathbb{R} \) is \( \mu^* \)-measurable if it holds for all sets \( E\subseteq\mathbb{R} \) that \( \mu^*(E) = \mu^*(E\cap A) + \mu^*(E\cap A^c) \).

First, we claim that if \( E\subseteq\mathbb{R} \) does not contain 0, then its outer measure is zero. Observe that E has cover \( \bigcup_{n=1}^{\infty}(-\infty,-1/n] \cup (0,\infty) \), demonstrating that E has outer measure zero. Next, we claim that if \(E\subseteq \mathbb{R} \) contains 0, then it has outer measure 1. Any cover of E by h-intervals contains some h-interval containing zero, which shows that the outer measure of E is at least 1. And \( \mathbb{R} \) is a cover of E, showing that the outer measure of E is at most 1.

It follows that for any sets A and E, we have \( \mu^*(E) = \mu^*(E\cap A) + \mu^*(E\cap A^c) \). Indeed, either E doesn't contain 0, in which case all the terms are zero, or E contains 0, in which case exactly one of \(E\cap A\) and \( E\cap A^c \) contains 0, and both sides of the equation are equal to 1. So, all sets A are \(\mu^*\)-measurable.

The other approach is to use completeness. We know that \( \mu^*\bigl(\{0\}^c\bigr) = 0 \), since \( \mu_F\bigl(\{0\}^c\bigr) = 0 \) and \(\mu_F\) is the restriction of \(\mu^*\) to the Borel σ-algebra. Carathéodory's Theorem says that the set of \(\mu^*\)-measurable sets is complete with respect to \( \mu^* \). Thus it contains all subsets of the \(\mu^*\)-null set \( \{0\}^c \). And since it also contains \( \{ 0 \} \), it contains all subsets of \(\mathbb{R}\).

I was going to try to draw the Cantor function on the board on Thursday, but I realized I should just link to its Wikipedia page, which has a better drawing then I'll make.

The Cantor function has the seemingly paradoxical property that at almost every x (i.e., for all x except those in Cantor set, which has measure zero), the function is locally constant at x, meaning there is an open interval around x on which the function is constant. And yet the function is not constant.

It's interesting to think about the measure \( \mu_f \) induced by the Cantor function f. I'll say a word or two about it on Thursday and we'll come to understand it better after we cover Chapter 3.

Update: I fixed a permissions problem with the class slides and now they should work.

Here is a video of today's lecture. And here is the contents of the board.

I changed problem 4 on the next homework assignment because we didn't get through Theorem 1.18 in class today (originally the problem I assigned was Folland 1.26, and your solution was supposed to use this theorem).

Last week I said I preferred for people to hand in their homework in person on paper, but I got quite a lot of homework submissions by email. So let me strengthen my statement and request that you please submit your homework in person on paper except in special circumstances (e.g., you will not be physically present in class for whatever reason). Please figure out how to print your homework at the Graduate Center; it can be done.

As I mentioned in class yesterday, the definition of monotone class on Homework #1 should not have included the requirement that it contains the base set X. This doesn't actually change the result of the problem, since every monotone class containing the algebra contains X, since the algebra contains X. But it does mess up the outlined proof in that the first step of it, proving that \( \mathcal{C}(E) \) is a monotone class, becomes much harder. On the other hand, I think everyone who did the problem using the outline ignored the issue entirely (or had an incorrect proof that \( \mathcal{C}(E) \) contained X, which I did not subtract any points for). But I'm sorry for messing it up.

Just to keep track of divergences from the textbook, note that the result that we stated in lecture yesterday (and will prove next week) comparing products of Borel σ-algebras to the Borel &sigma-algebra of a product space is different from the one in Folland. It's stated there (Proposition 1.5) that for separable metric spaces \( X_1,\ldots,X_n \) and \( X=X_1\times\cdots\times X_n \), we have \(\bigotimes_{j=1}^n\mathcal{B}_{X_j}=\mathcal{B}_X \). Our version of the result, with the same conclusion, assumes that \( X_j \) is a second countable topological space and allows for a countable product.

I've posted the first homework assignment, due in a week. Please hand in your assignment in person in class next Thursday. If you won't be in class, it's fine to email it to me instead, but if you will be there please just give me a physical copy.

Hello everyone and welcome to Math 70200, the second semester of real analysis. We meet in room 6417. We'll be covering Chapters 1, 2, 3, and 6 of Folland, building up measure theory and the Lebesgue integral. I'll stick reasonably closely to the textbook. See the calendar for more details. Here is the course syllabus.

Just about every mathematician uses the material in this course. In my field (probability), we also have our own special relationship with measures and σ-algebras, and they're more to us than foundational formalities; I'll try to get my perspective across to you when I can.

The key to success: spending lots and lots of time solving the homework problems.