# Interchanging limit measure and integral

I am reading a book and the autor skips a lot of step in a proof. I don't see if the next result can hold or not and how to prove or disprove it.

Let $$\{\mu_\epsilon\}_{\epsilon > 0}$$ be a sequence a positive finite measures on $$\mathbb{R}$$ and $$\mu$$ a positive finite measure on $$\mathbb{R}$$ such that for all A borel set of $$\mathbb{R}$$, $$\lim\limits_{\epsilon \to 0} \mu_\epsilon(A) =\mu(A)$$. Let $$f$$ be a bounded measurable function of $$L^\infty(\mathbb{R})$$. The author seems to use that $$\int_{\mathbb{R}} f(x) \mu(dx) = \lim\limits_{\epsilon \to 0} \int_{\mathbb{R}} f(x) \mu_{\epsilon}(dx)$$.

I tried to find results like that, but I didn't find any.

• are you sure that $\{\mu_\epsilon\}$ is a sequence' because $\epsilon$ seems real-valued, and so the map $\epsilon\mapsto\mu_\epsilon$ seems a real-valued function from $(0,\infty)$ to the space of Borel measures – Masacroso Jul 4 at 22:39
• You are right. I haven't see things with this point of view. But how can it help us? – jvc Jul 4 at 23:16

If $$\mu$$ is an infinite measure then $$\int f d\mu$$ may not make sense for bounded measurable $$f$$. So it is necessary to assume that $$\mu$$ is a finite measure. Any bounded measurable function is a uniform limit of simple functions. [ This is easy to see from the usual expression for simple functions that approximate the given function]. From the hypothesis we have $$\int f d\mu_{\epsilon} \to \int fd\mu$$ for simple functions $$f$$. So the result follows from triangle inequality if you use the fact that $$\mu (\mathbb R) <\infty$$ which implies $$\mu_{\epsilon} (\mathbb R)$$ remains bounded as $$\epsilon \to 0$$. [Note that it is enough to prove the result for each sequence $$(\epsilon_n)$$ tending to $$0$$].

• Thank you very much! – jvc Jul 4 at 23:37
• @Mars Plastic Thanks for correcting a silly mistake. – Kavi Rama Murthy Jul 4 at 23:39
• How do we know $\mu(\Bbb R)<\infty$? – David C. Ullrich Jul 4 at 23:57
• @DavidC.Ullrich Thanks for the comment. I read the question wrongly. However the claim is false if $\mu$ is not a finite measure. I have edited my answer. – Kavi Rama Murthy Jul 5 at 0:01
• But shouldn't the statement remain true for $\mu(\Bbb R)=\infty$ if we restrict it to only those $f\in L^\infty(\Bbb R)$ for which $\int f d\mu$ is well-defined, i.e. $f \in L^1(\mu)$ or $f\ge0$? – Mars Plastic Jul 5 at 0:11

A similar answer to the answer of master @KaviRamaMurthy. Note that we can decompose $$f=f^+-f^-$$ where

$$f^+:=\max\{f,0\},\qquad f^-:=\max\{-f,0\}\tag1$$

Then $$f^+,f^-:\Bbb R\to [0,\infty)$$, and there are sequences of $$\nu$$-simple functions such that $$\{s_n^+\}\uparrow f^+$$ and $$\{s^-_n\}\uparrow f^-$$ point-wise, and so (by the monotone convergence theorem) we have that

$$\int_{\Bbb R} f^+(x)\nu(dx)=\lim_{n\to\infty}\int_{\Bbb R} s^+_n(x)\nu(dx)\tag2$$

for any finite measure $$\nu$$ (a similar statement holds for $$f^-$$). Because the $$s_n^+$$ are non-negative $$\nu$$-simple functions we have that

$$\int_{\Bbb R}s^+_n(x)\nu(dx)=\sum_{k=1}^{m_n}c_{k,n}\nu(A_{k,n})\tag3$$

for some measurable sets $$A_{k,n}$$, some $$m_n\in\Bbb N$$ and some $$c_{k,n}\ge 0$$, and so

\begin{align}\int_{\Bbb R}s^+_n(x)\mu(dx)&=\sum_{k=1}^{m_n}c_{k,n}\mu(A_{k,n})\\ &=\sum_{k=1}^{m_n}c_{k,n}\lim_{\epsilon\to 0^+}\mu_\epsilon(A_{k,n})\\ &=\lim_{\epsilon\to 0^+}\sum_{k=1}^{m_n}c_{k,n}\mu_\epsilon(A_{k,n})\\ &=\lim_{\epsilon\to 0^+}\int_{\Bbb R}s^+_n(x)\mu_\epsilon(dx)\end{align}\tag4

Thus we want to show that

$$\int_{\Bbb R} f^+(x)\mu(dx)=\lim_{n\to\infty}\int_{\Bbb R} s^+_n(x)\mu(dx)=\lim_{n\to\infty}\lim_{\epsilon\to 0^+}\int_{\Bbb R}s^+_n(x)\mu_\epsilon(dx)\\=\lim_{\epsilon\to 0^+}\lim_{n\to\infty}\int_{\Bbb R}s^+_n(x)\mu_\epsilon(dx)= \lim_{\epsilon\to 0^+}\int_{\Bbb R}f^+(x)\mu_\epsilon(dx)\tag5$$

That is, we want to show that we can exchange the order of the limits in $$\rm (5)$$.

Now set $$I_{n,m}:=\int_{\Bbb R}s^+_n(x)\mu_{\epsilon_m}(dx)$$ for some arbitrary sequence $$\{\epsilon_m\}\downarrow 0$$. Then we have that

$$\lim_{m\to\infty}\sum_{k=0}^\infty\Delta_k I_{k,m}:=\lim_{m\to\infty}\lim_{n\to\infty}\sum_{k=0}^n(I_{k+1,m}-I_{k,m})\\=\lim_{m\to\infty}\lim_{n\to\infty}I_{n,m}\le \sup_m\mu_{\epsilon_m}(\Bbb R)\|f\|_\infty<\infty\tag6$$

Now note that the double sequence $$\{I_{n,m}\}$$ is non-negative and bounded, and so it is also $$\{\Delta_k I_{k,m}\}$$ because $$I_{n,m}$$ is increasing respect to $$n$$, so applying the dominated convergence theorem on $$\rm (6)$$ we find that

$$\lim_{m\to\infty}\lim_{n\to\infty} I_{n,m}=\lim_{m\to\infty}\sum_{k=0}^\infty\Delta_k I_{k,m}=\sum_{k=0}^\infty\lim_{m\to\infty}\Delta_k I_{k,m}=\lim_{n\to\infty}\lim_{m\to\infty} I_{n,m}$$

Thus it holds that

$$\int_{\Bbb R} f^+(x)\mu(dx)=\lim_{\epsilon\to 0^+}\int_{\Bbb R}f^+(x)\mu_\epsilon(dx)<\infty$$

Because a similar statement holds for $$f^-$$ we are done.