This is a problem that I've seen a few times on UCLA's analysis quals that I've been trying to solve, and I have a few questions.

Suppose $\mu_n,\mu$ are positive finite Borel measures on $\mathbb{R}$ such that $$ \int f\,d\mu_n \to \int f \,d\mu \qquad \forall f \in C_c(\mathbb{R}). $$ Show that $\limsup_{n\to\infty} \mu_n(K) \leq \mu(K)$ for all compact subsets $K\subset \mathbb{R}$.

Here's my attempt at a solution. Let $\Lambda: C_c(\mathbb{R}) \to \mathbb{R}$ be the functional given by $\Lambda(f) =\int f \,d\mu$. We have that $|\Lambda(f)| \leq ||f||_\infty \mu(\mathbb{R})$, which shows that $\Lambda$ is a bounded linear functional. By the Riesz Representation theorem, $\Lambda(f) = \int f \,d\tilde\mu$ where $\tilde \mu$ is a regular complex Borel measure.

Question 1: Can we assume/prove that $\mu = \tilde \mu$?

Suppose we can. Let $K\subset \mathbb{R}$ be a compact set. Fix $\epsilon>0$. As $\mu$ is regular, we can find an open set $U\supset K$ with $\mu(U\setminus K)<\epsilon$. Then, by Urysohn's lemma there exists $g$ continuous with $g(x)=0$ for $x\notin U$, $g(x)=1$ for $x\in K$, and $0\leq g(x)\leq 1$ for $x \in U\setminus K$. Clearly, $g$ has compact support. We have that $$ \int g \,d\mu_n \to \int_{U\setminus K} g \,d\mu + \int_K g\,d\mu \leq \mu(K) + \epsilon. $$ On the other hand, $$ \limsup_{n\to \infty} \mu_n(K)\leq \limsup_{n\to \infty}\left(\int_K g\,d\mu_n + \int_{U\setminus K} g\,d\mu_n \right)\leq \epsilon + \mu(K). $$ The first inequality follows since $\int_{U\setminus K}g\,d\mu_n \geq 0$, and the second follows by the above limit. As $\epsilon$ is arbitrary, we get the desired result.

Assuming the first question I made above, is this a valid solution? And, if question 1 is not true, what are some different methods for proving this? Also, my apologies if this question has been asked before; I couldn't seem to find it anywhere on the site.

Edit: I was able to solve the problem without needing to know whether or not Question 1 is true. Regularity of $\mu$ may be already true, but my answer below shows how to get the open set $U$ in my argument above without knowing that $\mu$ is regular.

  • $\begingroup$ Where do you use Q1 in your proof? $\endgroup$ – Giovanni Sep 20 '15 at 17:09
  • $\begingroup$ I use it to assume that $\mu $ is regular, in order to choose $U $ add I have. $\endgroup$ – Andrew Sep 20 '15 at 17:10
  • $\begingroup$ It is a Borel measure by assumption, this should imply that it is regular, doesn't it? $\endgroup$ – Giovanni Sep 20 '15 at 17:11
  • $\begingroup$ Maybe some assume Borel means regular by comvention, but I do not. It may be that this problem assumes this convention, and I just don't know. $\endgroup$ – Andrew Sep 20 '15 at 17:14
  • $\begingroup$ Ok, here is the thing: you can prove that any finite Borel measure on a compact space is regular. You can use this in this context extending this result by sigma finiteness to $\mathbb{R}$ (which can be expressed as union of compact sets), hence you can assume that $\mu$ is regular. $\endgroup$ – Giovanni Sep 20 '15 at 17:22

Define $f_m\in C_c(\mathbb{R})$ by $f_m(x)=\left(1-m d(x,K)\right)_+$ where $d(x,K)=\inf(d(x,k): k\in K)$ is the distance of $x$ from $K$. Then $f_m\geq 1_K$ so that $$\int f_m \,d\mu_n\geq \int 1_K \,d\mu_n.\tag1$$ Taking the $\limsup_n$ in (1) gives $$\int f_m \,d\mu\geq \limsup_n \mu_n(K).\tag2$$ Letting $m\to\infty$ in (2) gives the result.

  • $\begingroup$ Definitely not the first proof I would have thought of, but it is very nice! A quick question: those $f_m$ are dominated by $f_1$, correct? Then dominated convergence lets you compute the $m\to \infty$ limit? $\endgroup$ – Andrew Sep 20 '15 at 23:21
  • $\begingroup$ @Andrew Correct. $(f_m)$ is a decreasing sequence of functions. $\endgroup$ – user940 Sep 21 '15 at 2:25

With help from Giovanni, I was able to figure out the correct thing to search, and from this question I was able to find a different way to avoid my Q1 altogether.

We claim that $K=\bigcap_{n=1}^\infty V_n$ for some open sets $V_n$. Indeed, fix a positive integer $n$. Then, consider the sets $\{ B(y,1/n)\}_{y\in K}$. By compactness of $K$ we get a finite subcover, and so we can define $V_n$ to be the union of this finite subcover. We see that $K\subset \bigcap_{n=1}^\infty V_n$. Now, if $x\in V_n$ for each $n$, then we have a sequence $(y_n)$ with $y_n \in K$ and $|x-y_n|<1/n$. It follows that $y_n \to x$ as $n\to \infty$, so $x \in K$ as $K$ is closed. This proves the claim.

Finally, the last step I needed in my proof was to prove the existence of the set $U\supset K$ with $\mu(U\setminus K)<\epsilon$. Well, for given $\epsilon>0$, continuity from above implies that $$ \mu(K) = \mu\left(\bigcap_{n=1}^\infty V_n\right) = \lim_{n\to \infty} \mu\left(\bigcap_{k=1}^n V_k\right) $$ (which is justified as $\mu(V_1)<\infty$; it is just a finite union of bounded open sets). Hence, the desired $U=\bigcap_{k=1}^n V_k$ for large enough $n$ because $\mu(V_n\setminus K)=\mu(V_n) - \mu(K) < \epsilon$.

The proof resumes as in the original post after Question 1.


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