Analysis Problem of the Day 90

Today’s problem appeared as Problem 2 on the UCLA Spring 2019 Analysis Qual:

Problem 2. Let $\mu_n,\mu$ be Borel probability measures on $[0,1]$ such that $\mu$ is atomless, i.e. $\mu(\{t\})=0$ for all $t \in [0,1],$ and suppose $\mu_n \overset{*}{\rightharpoonup} \mu$ as measures. Define $F_n(t):=\mu_n([0,t])$ and $F(t):=\mu([0,t]).$ Show that $F_n \to F$ uniformly on $[0,1].$

Solution: Note that $F_n,F$ are bounded monotonically increasing functions with $F_n(0)=F(0)=0,F_n(1)=F(1)=1.$ Since $\mu$ has no atoms, one has that $F$ has no jump discontinuities, and since a monotonic function can only have jump discontinuities, we conclude that $F$ is continuous on $[0,1].$ By a classic result from the theory of monotone functions, if a sequence of monotone functions converges pointwise to a continuous monotone function, then the convergence is uniform (which we quickly prove below as a lemma). Supposing this, it suffices to show that $F_n(t) \to F(t)$ pointwise. But indeed, for any $t \in [0,1],$ define $f_k^t:[0,1] \to [0,1]$ to be $f_k(x) \equiv 1$ on $[0,t-\frac{1}{k}], f_k(x) \equiv 0$ on $[t,1],$ and $f_k(x)=1-k(x-t+\frac{1}{k})$ on $[t-\frac{1}{k},t].$ Then, $f_k$ is clearly continuous and $f_k$ increases monotonically to $\chi_{[0,t]}(x).$ It follows that since $\mu_n \to \mu$ in the weak $-*$ topology that $\int f_k \mu_n \to \int f_k d\mu$ as $n \to \infty.$ Now, by the Monotone Convergence Theorem, $\int f_k d\mu_n$ increases to $\mu_n([0,t])$ and $\int f_k d\mu$ increases to $\mu([0,t])$ as $k \to \infty.$ Thus, if for sake of contradiction, say, $\mu_{n_j}([0,t]) \leq \mu([0,t])-\epsilon$ for some $t \in [0,1]$ and infinitely many $n_j,$ then since

$\int f_k d\mu_n \leq \int f_{K} d\mu_n \leq \mu_n([0,t]), \quad \int f_k d\mu \leq \int f_K d\mu \leq \mu([0,t])$

for $K \geq k,$ picking $K$ large enough such that $\mu([0,t])-\int f_K d\mu < \frac{\epsilon}{2},$ we get that $\int f_{K} d\mu_{n_j} \leq \mu([0,t])-\epsilon$ for all $n_j,$ which is a contradiction since $\int f_K d(\mu_{n_j} -\mu) \to 0$ as $j \to \infty.$ Thus, we conclude that $\mu_n([0,t]) \to \mu([0,t])$ pointwise, and therefore by the above discussion, $F_n(t) \to F(t)$ uniformly.

Proof of lemma: Suppose $f_n,f$ be increasing functions on $[0,1],$ and suppose $f_n \to f$ pointwise and $f$ is continuous. Then, for any $\epsilon >0,$ pick $\delta>0$ such that $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$ and pick a finite set of points $x_0<x_1<...<x_m$ such that any $x \in [0,1]$ satisfies $|x-x_i|<\delta$ for some $i.$ By pointwise convergence, let $n$ be large enough such that $|f_n(x_i)-f(x_i)|<\epsilon$ for all $i.$ Then, for any $x \in [x_i,x_{i+1}],$

$|f(x)-f_n(x)| \leq |f(x)-f(x_i)| + |f(x_i) - f_n(x_i)| +|f_n(x_i)-f_n(x)|$

$\leq 2\epsilon+|f_n(x_{i+1})-f_n(x_i)|\leq 2\epsilon + |f_n(x_{i+1}) -f(x_{i+1})|$

$+|f(x_{i+1})-f(x_i)|+|f(x_i)-f_n(x_i)| < 5\epsilon.$

Thus, $|f(x)-f_n(x)|<\epsilon$ for any $x \in [0,1],$ i.e. $f_n \to f$ uniformly.

Remark: The statement need not hold if $\mu$ has an atom. Take, say, $\mu$ be the Dirac delta at $x=1,$ and let $\mu_n(A):=\int_A nx^{n-1} dx.$ Then, clearly, $F_n$ does not converge to $F$ uniformly, even though $\mu_n \to \mu,$ i.e. $\int_0^1 nx^{n-1} f(x) dx \to f(1)$ for all $f \in C([0,1]).$

Leave a Reply Cancel reply