# Lebesgue integral over a collection of sets

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