# Thread: [SOLVED] Limit of integral on measurable sets

1. ## [SOLVED] Limit of integral on measurable sets

I'm struggling with a prelim practice problem and need some help; note that E may have infinite measure.

Claim: Let $\displaystyle f_n$ be a sequence of integrable functions such that $\displaystyle f_n \rightarrow f$ a.e. Assume that $\displaystyle \lim_{n \rightarrow \infty}{\int{f_n}} = \int{f} < \infty$. Prove or disprove that for any measurable set E, $\displaystyle \lim_{n \rightarrow \infty}{\int_{E}{f_n}} = \int_{E}{f}$.

I't's not clear that the conjecture is even true but I haven't been able to find a counter example if it's not. If $\displaystyle f_n$ and $\displaystyle f$ are non-negative, you can use Fatou's Lemma to prove the result. If we can show that $\displaystyle \int{f_n^+} \rightarrow \int{f^+}$ then we can use the generalized Lebesque Dom Conv Thm. but I haven't been able to prove the limit. I also tried assuming that E has finite measure and applying Egoroff's theorem to no avail. I also tried assuming that the claim is true for E of finite measure and then proving the more general case - didn't get home on that approach either. I also played a little with signed measure ideas (positive and negative sets) but that just seemed to make the problem more abstract without adding any value. The way through may be in these somewhere in these ideas but I haven't been able to pull it out.

2. For E with finite measure I think you can just use
$\displaystyle |\int_Ef_n-f|\le\int_E\epsilon\le\mu(E)\cdot\epsilon$.

For the general case I would look at something like

$\displaystyle I_Ef_n\to I_E f$ where $\displaystyle I_E(x)=1$ if $\displaystyle x\in E$, and then show that $\displaystyle \int I_Ef_n\to\int I_E f$

3. Originally Posted by huram2215
I'm struggling with a prelim practice problem and need some help; note that E may have infinite measure.

Claim: Let $\displaystyle f_n$ be a sequence of integrable functions such that $\displaystyle f_n \rightarrow f$ a.e. Assume that $\displaystyle \lim_{n \rightarrow \infty}{\int{f_n}} = \int{f} < \infty$. Prove or disprove that for any measurable set E, $\displaystyle \lim_{n \rightarrow \infty}{\int_{E}{f_n}} = \int_{E}{f}$.
This remind's me Scheffé's lemma. The positiveness assumption (or the absolute values) involved in Scheffé's lemma suggests that the results fails otherwise, but I can't think of a counterexample either. I will give it further thoughts...

4. Putnam120: For the case of E finite, you idea works if $\displaystyle f_n \rightarrow f$ uniformly. Hence the potential for using Egoroff's Theorem. However that still leaves the small set to deal with where $\displaystyle f_n$ may not converge to $\displaystyle f$ uniformly.

5. Originally Posted by Laurent
This remind's me Scheffé's lemma. The positiveness assumption (or the absolute values) involved in Scheffé's lemma suggests that the results fails otherwise, but I can't think of a counterexample either. I will give it further thoughts...
After further thoughts...

I considered the well-known counterexample to the bounded convergence theorem, namely the "traveling bump" (I don't know the English name) : define the sequence of functions $\displaystyle \phi_n(t)= {\bf 1}_{(n\leq t<n+1)}$ (it equals 1 if $\displaystyle t\in[n,n+1)$ and 0 otherwise). Any other smoother bump would do fine. We have $\displaystyle \int\phi_n(t) dt = 1$ for any $\displaystyle n$, but $\displaystyle \phi_n(t)\to_n 0$ for every $\displaystyle t$. Of course, the convergence to 0 is not uniform and can't be dominated either.

Then I defined $\displaystyle f_n(t)=\phi_n(t)-\phi_n(-t)$. This time, two opposite bumps travel toward $\displaystyle \pm\infty$. We have $\displaystyle f_n(t)\to_n 0$ for all $\displaystyle t$, and $\displaystyle \int f_n(t)dt = 0$. This is the situation of the question. However, $\displaystyle \int_{\mathbb{R}_+} f_n(t)dt = 1$ for every $\displaystyle n$.

6. Laurent: Very nice counter example. The deduction that you applied (including the introduction to Scheffe's Lemma) gives good incite into how you approached the problem. Thank you.

7. Originally Posted by huram2215
Putnam120: For the case of E finite, you idea works if $\displaystyle f_n \rightarrow f$ uniformly. Hence the potential for using Egoroff's Theorem. However that still leaves the small set to deal with where $\displaystyle f_n$ may not converge to $\displaystyle f$ uniformly.
Thanks, I only noticed that after I woke up the next day. Guess that teaches me never to do analysis when I'm tired.