Another question in my studies (of Lebesgue integration):
I'm given a continuous function $f:\mathbb{R}\to\mathbb{R}$, a nonempty compact set $E\subseteq\mathbb{R}$, and a sequence of nonempty compact sets $E_i$ such that $\lim d(E_i , E) = 0$where this $d$ is Hausdorff distance.
I am asked to prove that $\lim \int_{E_i} f=\int_E f$
Visually, this seems pretty straightforward, the $E_i$ are being forced to look more and more like $E$, and the continuity of $f$ ensures that the farther along the $E_i$ we pick, $f$ has to vary less and less where $E_i$ and $E$ don't coincide.
I've been at this for a while and seem to be stuck. For a while though I hadn't noticed the compactness assumed of $E$ and the $E_i$. Putting the continuity and convergence in Hausdorff distance together, I see that for any $x\in E$, $\varepsilon>0$, and $\delta > 0$, there is eventually some $E_i$ for which I can find a $y\in E_i$ within $\delta$ of $x$ and with $|f(x)-f(y)|<\varepsilon$. I guess using compactness, I can make this uniform over all $x\in E$ (so that my picture described above is actually as nicely behaved as I was probably picturing it to begin with).
At this point though I'm not sure how to proceed. Should I try to bound the integral over the symmetric difference of $E_i$ and $E$ or something like that? I'm not sure how I would control it, despite knowing it would be constrained somehow by "nearby" values of $f$. Should I cut up $E$ somehow, or select some dense subset of $E$ to use representative values of $f$ on? Is there some subtlety my visual intuition is overlooking?
Added
It seems like there is a counterexample to this theorem as stated:
Let $E=[0,1]$ and let $E_i = \{ k 2^{-i} : k=0,1,\dots,2^i \}$ so that $\array{ E_1 & = & \{0,\frac{1}{2}, 1\} \\ E_2 & = & \{0,\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\} \\ & \vdots }$
Then, for example, with $f$ taken to be the constant $1$ function on $[0,1]$, we have $\int_{E_i} f = 0$ for all $i$, yet $\int_{E} f = 1$.