Csdrhrt

Pleas check my proof of the following assertion.

Let ${E_k}$ be a sequence of measurable sets, and $E_k nearrow E$. Suppose $f(x)$ is nonnegative and measurable on $E$. Prove

$$int_E f(x) , dx = lim_{k to infty} int_{E_k} f(x) , dx$$

Proof. First note that

$$int_{E_k} f(x) , dx= int_E f(x)chi_{E_k}(x) dx$$

and we have (using our hypothesis in the second equality)

$$lim f(x)chi_{E_k}(x) = f(x) lim chi_{E_k}(x) stackrel{tiny{HYP.}}{=} f(x)chi_E(x) tag{?}$$

Furthermore, $left(f(x)chi_{E_k}(x)right)$ is monotone increasing, nonnegative, and measurable (product of two measurable functions). Therefore, by the MCT

$$lim_k int_{E_k} f(x) , dx= lim_k int_E f(x)chi_{E_k}(x) , dx stackrel{tiny{MCT}}= int_E f(x) , dx$$

So the proof is complete.

I took for granted in $(?)$ that $lim chi_{E_k}(x) = chi_E(x)$. So to prove that we need to show that for all $epsilon > 0$, and any $x in E$, there exists and $N$, such that whenever $k > N$ we have

$$ |chi_E(x) - chi_{E_k}(x)| < epsilon$$

But this can be relaxed to almost any $x in E$ for the MCT. So, let $epsilon > 0$ and let $x in E$. The set of $x$ on the boundary of of $E$ has measure zero, since that set has side length zero in at least one dimension (??). If $x$ is in the interior of $E$ we can, by hypothesis, produce an $E_N subset E$ such that $x in E_N = E cap E_N$. Therefore, for all $k > N$ we have $x in E_k$ which means that $chi_E(x) = 1$ and $chi_{E_k}(x) = 1$, which clearly satisfies the inequality above.

Something about that doesn't sit right with me, the (??) part mostly. I was thinking about setting this problem up a different way too.

Not sure if this answers your question, but we can show that if $E_k uparrow E$ then we have that:
$$
muleft(bigcup_{k = 1}^{infty}E_kright) = lim_{n to infty} muleft(bigcup_{k = 1}^{n}E_kright)
$$
This is done by defining $G_1 = E_1$, $G_2 = E_2 setminus E_1$, $G_3 = E_3 setminus E_2$, etc, with $G_ k = E_k setminus E_{k-1}$. Then, we see that these sets are disjoint, and $bigcup G_k = E = bigcup E_k$. By disjoint additivity, we have:
$$
mu(E) = sum_{k = 1}^infty mu(G_k) = lim_{n to infty} sum_{k = 1}^{n}mu = (G_k) = lim_{n to infty} mu left(bigcup_{k = 1}^{n}G_kright) = lim_{n to infty} mu left(bigcup_{k = 1}^{n}E_kright)
$$
as we needed. Having proven the two sets have the same measure, it follows that the indicator function converges accordingly (in an A.E sense)