Lebesgue integral over a collection of sets

Question

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.

Answer 1

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.

Answer 2

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.