Let $E$ and $\langle E_n \rangle$ be measurable sets in $\mathbb{R}$. Suppose that $f$ is Lebesgue integrable over $E$. If $E_n\subset E$ for all $n$ and $\displaystyle \lim_{n\to \infty} m(E_n)=m(E)<+\infty$, show that $\displaystyle \lim_{n\to \infty} \int_{E_n} f(x)\ dx= \int_E f(x)\ dx.$

The theorems at my disposal are pretty much any theorem from the first 4 chapters of Royden 3rd Edition. The technique I tried involved creating an increasing sequence of sets (call them $D_k$) from the $E_n$'s via unions, then create a sequence of functions $f_k=f \chi_{D_k}$ however that will only converge if $\bigcup D_k=E$ which I am not sure it will. Can anyone give me any pointers to help with this method? Or if I'm on the wrong track help steer me on the correct track.

  • Can you not define functions $f_n$ which agree with $f$ on $E_n$ and are $0$ everywhere else and then use the dominated convergence theorem? – James Jun 6 '13 at 20:42

EDIT: Thanks to Ilya's comments, I realized that my original proof relied on an assertion that was totally wrong. I thought about it for a bit and came up with this one instead. I think it's a bit more straightforward.

In this problem we are integrating with respect to Lebesgue measure, which is $\sigma$-finite. Therefore to invoke dominated convergence, we do not need to show convergence a.e. Rather, we need only show: $$ f \cdot \mathbb{1}_{E_n} \;\to\; f \cdot \mathbb{1}_{E} \quad \textrm{in measure} $$ To see this, note that for any $\epsilon>0$, $$ \lim_{n \to \infty} m \Big\{x \in E: \big| \; f \cdot \mathbb{1}_{E_n} - f \cdot \mathbb{1}_{E} \; \big| > \epsilon \Big\} = \lim_{n \to \infty} m(E-E_n)$$ Which is 0 by the assumption that $\lim_{n \to \infty} m(E_n) = m(E)$. Therefore $f \cdot \mathbb{1}_{E_n} \to f \cdot \mathbb{1}_{E}$ in measure and applying dominated convergence gives the result.

  • How do you show that $\lim \mu(E - E_n) = 0$ implies $\lim \int_{E - E_n}f\mathrm dm = 0$ as $f$ is not necessarily bounded? It seems to be the main part of the question. – Ilya Jun 6 '13 at 16:38
  • Ilya: You're right that it is the main part of the question! I was hoping to still leave the bulk of the work out there and just recast the problem, but then I suppose you can't really call it an "answer." I'll edit to make the argument explicit. Thanks! – gogurt Jun 6 '13 at 17:41
  • The point is that I believe OP was able to recast the problem (the way you did) as well. Anyways, I'm looking forward to your edit, since my solution may be a kinda overkill. – Ilya Jun 6 '13 at 18:05

Perhaps, there is a simple proof for this fact, but anyways. Let us define a new measure $n$ by $$ n(\mathrm dx) := f(x)m(\mathrm dx) \tag{1} $$ and since $f$ is $m$-integrable over $E$, we have that $n$ is finite on $E$. We need to show that $$ \lim_k m(E_k)= m(E)\implies \lim_k n(E_k)= n(E). $$ From $(1)$ we have that $n\ll m$, so that we can use an $\varepsilon$-$\delta$ definition of the absolute continuity (see e.g. the 1st paragraph here). Fix $\varepsilon>0$, and pick up $\delta>0$ such that $|m(A)|<\delta$ implies $|n(A)|<\varepsilon$. Furthermore, there exists $K$ such that $|m(E_k)|<\delta$ for all $k>K$. As a result, for any $\epsilon>0$ there exists $K$ such that $|n(E_k)|<\varepsilon$ for all $k>K$ and thus the convergence holds true.

  • After doing this how would I relate that equality back with $\int_{E_n} f(x)\ dx$? – Richard Jun 6 '13 at 0:27
  • @Richard: I've edited the proof - you may wanna take a look. – Ilya Jun 6 '13 at 13:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.