Here are some hints:
To see why $E^{x_1}$ is borel, consider the (obviously borel) function
$f(x_2) = (x_1,x_2)$. Do you see why $E^{x_1}$ is the preimage of a borel set by a borel function?
Remember that the lebesgue measurable sets are exactly
$\{ \text{borel sets} \} \cup \{ \text{extra null sets} \}$, so we will probably have to use these extra null sets somehow.
More concretely: Take your favorite (lebesgue) nonmeasurable set $X$. Can you show that $X \times \{0\}$ is (lebesgue) measurable in $\mathbb{R}^2$? Remember you'll probably have to use nullsets!
Once you've done this, $(X \times \{0\})^0$ is obviously $X$, which was nonmeasurable by assumption.
I hope this helps ^_^
You need to be careful about how exactly your subsequences are constructed.
Take a sequence $r_k\nearrow 1$ as $k\to\infty$, and let $B_k := B(0,r_k)$.
Let us start with $B_1$. By the assumptions there is a subsequence $f_{\sigma(1,1)}, f_{\sigma(1,2)}, f_{\sigma(1,3)},\dotsc$ which converges pointwise a.e. in $B_1$. Let $Z_1\subset B_1$ be the measure zero set where the pointwise convergence fails, so $(f_{\sigma(1,n)})$ converges pointwise everywhere in $B_1\setminus Z_1$ as $n\to\infty$.
Now, for $B_2$, we let $(f_{\sigma(2,n)})$ be a further subsequence of $(f_{\sigma(1,n)})$ that converges pointwise a.e. in $B_2$. This is possible since $(f_{\sigma(1,n)})$ certainly converges in $L_{\rm loc}^p$ as well. Continuing in this manner, we let $(f_{\sigma(k,n)})$ be a subsequence of the $(f_n)$ with the properties that
- $(f_{\sigma(k,n)})$ is a subsequence of $(f_{\sigma(k-1,n)})$
- $(f_{\sigma(k,n)})$ converges pointwise a.e. in $B_k$, with exceptional measure zero set $Z_k\subset B_k$
Let's arrange everything into a grid:
\begin{equation*}
\begin{array}{cccc}
f_{\sigma(1,1)} & f_{\sigma(1,2)} & f_{\sigma(1,3)} &\cdots \\
f_{\sigma(2,1)} &f_{\sigma(2,2)} & f_{\sigma(2,3)} & \cdots \\
f_{\sigma(3,1)} &f_{\sigma(3,2)} & f_{\sigma(3,3)} & \cdots \\
\vdots & \vdots & \vdots & \ddots
\end{array}
\end{equation*}
It's then not hard to see that the diagonal sequence $g_n = f_{\sigma(n,n)}$ is such that $(g_n)_{n\geq k}$ is a subsequence of $(f_{\sigma(k,n)})_{n\geq 1}$.
Let $Z = \bigcup_i Z_i$, and note that $Z$ is measure zero as the countable union of measure zero sets. Let $x\in B\setminus Z$. Then $x\in B_k\setminus Z_k$ for some $k$. For this $k$, we see that $g_n(x)\to f(x)$ as $n\to\infty$ since, for $n\geq k$, $g_n(x) = f_{\sigma(n,n)}(x)$ is a subsequence of $(f_{\sigma(k,n)}(x))$, the latter of which has limit $f(x)$ by construction. Thus the $(g_n)$ converge pointwise everywhere in $B\setminus Z$, and is thus the desired subsequence of the original $(f_n)$.
Best Answer
The basic approximation theorem of Measure Theory (Ref. Theorem D, p. 56 of Halmos' Measure Theory) shows that given any Borel set $B$ in $[0,1]$ there exists a seqeuence of sets $(B_n)$ such that each $B_n$ is a finite disjoint union of intervals and $m (B\Delta B_n) \to 0$. ($m$ denotes Lebesgue measure). By further approximation we may take these intervals to have rational end points. You can now finish the proof using the fact that $|\int_B f_{k_n} -\int_{B_m} f_{k_n}| \leq C m(B\Delta B_m)$ where $C$ is a bound for $f_n$'s.