3

Given that $f$ is an integrable function on $X$ and $\{E_k\}_{k=1}^\infty$ where each $E_k$ is a measurable set such that $\lim_{k\rightarrow \infty} \mu(E_k) = 0$

Can we show that $$\lim_{k\rightarrow \infty} \int_{E_k} fd\mu = 0$$

I want to prove like this: $$|\int_{E_k} fd\mu| \leq sup|f|\cdot \mu(E_k) \rightarrow 0 $$ The problem is when $|f| \rightarrow \infty$, I'm not sure if this is valid.

And if we remove the condition $f$ integrable and instead make f positive measurable, does the result still hold?

blancmange
  • 302

2 Answers2

3

Let $f_k = f\chi_{E_k}$. Note $f_k \to 0$ almost everywhere since $\mu(E_k) \to 0$. Also, $|f_k| \le |f|$ which is integrable, so Dominated Convergence Theorem gives you the result.

mathworker21
  • 35,247
  • 1
  • 34
  • 88
  • If we weaken the condition f integrable to f measurable do we still have the result? – blancmange Mar 31 '17 at 09:04
  • 2
    No. $f$ could blow off to infinity quickly on a set of small measure. Consider $\frac{1}{x}$ on $(0,1]$. – mathworker21 Mar 31 '17 at 09:10
  • I think you are using $\chi_{E_k}$ something like $\mu(E_k)$. Otherwise it won't tend to zero. But then $f_k$ is not bounded by f. – quallenjäger Mar 31 '17 at 09:31
  • I am using $\chi_{E_k}$ correctly. Note $f_k(x) = 0$ for $x \not \in E_k$. Since $\mu(E_k) \to 0$, $f_k$ tends to $0$ for a.e. $x$. – mathworker21 Mar 31 '17 at 09:35
  • Sorry I didn't see you mentioned almost everywhere in the convergence. – quallenjäger Mar 31 '17 at 09:59
  • 2
    @mathworker21 Let $\lambda $ be the Lebesgue measure on $I=[0,1]$. Let $m\geq 1$. We can write $m$ in a unique manner $m=2^n+j$ with $0\leq j<2^n$. Define the subset $E_m$ to be $\displaystyle E_m=[\frac{j}{2^n}, \frac{j+1}{2^n}]$. Then it is easy to see that $\lambda(E_m)\to 0$ as $m\to \infty$. But let $x\in ]0,1[$; for $n\geq 1$, there exists a $j$ such that $j\leq 2^nx<j+1$. For $m=2^n+j$, this give $x\in E_m$; hence there is an infinite number of $m$ such that $\chi_{E_m}(x)=1$. As a consequence, $\chi_{E_m}$ does not go to $0$ almost everywhere. – Kelenner Mar 31 '17 at 11:27
  • I'm confused. How can $x$ be in $E_m$ if $E_m \subseteq [0,1]$ and $x \not \in [0,1]$? – mathworker21 Mar 31 '17 at 11:51
  • Sorry, I do not understand why you write $x\not \in [0,1]$ ? – Kelenner Mar 31 '17 at 11:54
  • Kelenner you are right – blancmange Mar 31 '17 at 13:20
  • 3
    @blancmange I think that you can repair mathworker21's solution in the following way. Suppose that $\int |f|\chi_{E_k}d\mu$ does not go to $0$. Then we can find a $q>0$ and a subsequence $E_{k_j}$ such that $\int |f|\chi_{E_{k_j}}d\mu \geq q>0$ for all $j$. Now the sequence $\chi_{E_{k_j}}\to 0$ in $L^1$. It is well known that in this situation there is a subsequence $\chi_{E_{k_{j_l}}}$ that goes to $0$ almost everywhere. Now mathworker's solution give that $\int |f|\chi_{E_{k_{j_l}}}d\mu \to 0$ as $l\to \infty$, a contradiction. – Kelenner Mar 31 '17 at 13:58
  • @Kelenner, sorry I was being silly. I misunderstood the notation $]0,1[$ but should've been able to figure out what you meant from context. And yes, nice fix! Sorry again about my mistake. – mathworker21 Mar 31 '17 at 18:12
  • I think my mistake has a probabilistic interpretation via Borel Cantelli. There can be a sequence of events with probabilities going to $0$, but the probability infinitely many of them happen is $1$. – mathworker21 Mar 31 '17 at 18:13
1

Write integral over $E_k$ as integral of f*indicator ($E_k$) = $f_k$ , it should work

RiezFrechetKolmogorov
  • 238
  • No it is just a multiplication actually... If you cant see why it works you can consider $f$ as a simple function and extend the result to all integrable $f$. – RiezFrechetKolmogorov Mar 31 '17 at 09:03