Math 70200

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.