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.

Thank you for your help.

  • $\begingroup$ 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 $\endgroup$ – Masacroso Jul 4 at 22:39
  • $\begingroup$ You are right. I haven't see things with this point of view. But how can it help us? $\endgroup$ – 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 $0$].

  • $\begingroup$ Thank you very much! $\endgroup$ – jvc Jul 4 at 23:37
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    $\begingroup$ @Mars Plastic Thanks for correcting a silly mistake. $\endgroup$ – Kavi Rama Murthy Jul 4 at 23:39
  • $\begingroup$ How do we know $\mu(\Bbb R)<\infty$? $\endgroup$ – David C. Ullrich Jul 4 at 23:57
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Jul 5 at 0:01
  • $\begingroup$ 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 $f\ge0$? $\endgroup$ – 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.


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