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Let $\mu: \mathbb{B}(\mathbb{R}) \to [0, \infty[$ be a finite Borel-measure. Show that there exists a sequence of Borel-measures $\{\mu_n\}_{n=1}^\infty$ with finite supports such that

$$\mu_n \stackrel{v}{\to} \mu, \quad\sup_{n=1}^\infty \mu_n(\mathbb{R}) < \infty$$

Here $\mu_n \stackrel{v}{\to} \mu$ means vague convergence. This means that for every $a,b \in \mathbb{R}$ with $\mu\{a\}= \mu\{b\}=0$ we have

$$\mu_n(]a,b])\stackrel{n \to \infty}\to \mu(]a,b])$$

Equivalently,

$$\int fd\mu_n \to \int fd \mu$$

for all continuous compactly supported functions $f: \mathbb{R} \to \mathbb{R}$.

Attempt:

I am hinted to use the sequence

$$\mu_n = \sum_{j=-n2^n +1}^{n2^n}\mu\left(\left]\frac{j-1}{2^n}, \frac{j}{2^n}\right]\right) \delta_{\{j/2^n\}}$$

where $\delta_{a}$ is the Dirac measure at $a \in \mathbb{R}$.

I'm a bit unsure how to prove this. Using the definition seems a little bit tedious. Maybe I can associate the distribution functions

$$F_n(x) = \mu_n(]-\infty, x]), \quad F(x) = \mu(]-\infty, x]), x \in \mathbb{R}$$

and show that for $x \notin D(F)$ (the points where $F$ is not continuous, or equivalently where the $\mu$-measure of the singelton is non-zero) we have

$$F_n(x) \to F(x)$$

Any help into the right direction is appreciated!

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  • $\begingroup$ I would suggest to use the characterization in terms of compactly supported continuous functions. This should be easier than using the distribution function. Use that by uniform continuity, $f$ is "almost constant" on each interval $](j-1)/2^n, j/2^n]$, for $n$ large. $\endgroup$ – PhoemueX Dec 12 '19 at 20:49
  • $\begingroup$ I see that $f$ is uniformly continuous since it is compactly supported. But can you elaborate what you mean with "almost constant"? And how that helps to show the integrals converges to what we want? I tried this earlier but didn't get anywhere. Maybe post an answer? $\endgroup$ – user661541 Dec 12 '19 at 20:51
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I will use the characterization in terms of compactly supported functions that you mention.

Let $\epsilon > 0$. Since $f$ is uniformly continuous, there is $\delta > 0$ such that $|f(x) - f(y)| \leq \epsilon / (1 + \mu(\Bbb{R}))$ for $|x-y| \leq \delta$. Now, choose $n_0 = n_0(f, \delta)$ so large that $2^{-n_0} \leq \delta$ and $\mathrm{supp} f \subset (-n_0,n_0]$, and let $n \geq n_0$ be arbitrary.

If you define $F := \sum_{j = -n 2^n + 1}^{n 2^n} 1_{((j-1)/2^n, j/2^n]} f(j/2^n)$, then $|F(x) - f(x)| \leq \epsilon$ for all $x$. Indeed, if $x \notin (-n,n] \supset (-n_0,n_0]$, then $F(x) = f(x) = 0$. Otherwise, we have $x \in ((j-1)/2^n, j/2^n]$ for a unique $j \in \{-n2^n + 1,\dots,n2^n\}$, and hence $|F(x) - f(x)| = |f(j/2^n) - f(x)| \leq \epsilon / (1 + \mu(\Bbb{R}))$, since $|x - j/2^n| \leq 2^{-n} \leq \delta$.

Finally, note that $\int F \, d \mu = \int f \, d \mu_n$ (why?!).

Therefore $$ \bigg| \int f d \mu_n - \int f d \mu \bigg| = \bigg| \int F d \mu - \int f d \mu \bigg| \leq \int |F - f| d \mu \leq \frac{\epsilon}{1 + \mu(\Bbb{R})} \cdot \int 1 d \mu \leq \epsilon, $$ for all $n \geq n_0$.

Since this holds for any compactly supported, continuous $f$, we are done.

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  • $\begingroup$ Thanks so much! I will look at it! For the why question: I guess this is because we have: $\int f d(\mu_1 + \mu_2) = \int f d \mu_1 + \int f d \mu_2$ and because the integral over a dirac measure is evaluating at a point? $\endgroup$ – user661541 Dec 12 '19 at 21:05
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    $\begingroup$ @user661541: Yes, exactly. Also, $\int 1_A \, d \mu = \mu(A)$. $\endgroup$ – PhoemueX Dec 12 '19 at 21:09
  • $\begingroup$ Your last line contains a little mistake. $\int 1 d \mu = \mu(\mathbb{R})$. It is not given that this is a $p$-measure. Fortunately, this doesn't make the proof unvalid. Thanks for the help it is very clear! $\endgroup$ – user661541 Dec 13 '19 at 9:40
  • $\begingroup$ Moreover, you show the estimate for one $n$ sufficiently large. We need it for all $n$ sufficiently large but this is also easily fixed. $\endgroup$ – user661541 Dec 13 '19 at 9:50
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    $\begingroup$ @user661541: Thanks for you comments. I incorporated them into my updated answer. Before you ask: I am using $1 + \mu(\Bbb{R})$ instead of $\mu(\Bbb{R})$ to avoid dividing by zero in case of $\mu(\Bbb{R}) = 0$ :) $\endgroup$ – PhoemueX Dec 13 '19 at 10:59

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