Property of functions: do there exist counterexamples? I was assigned the following:

Let $\tau_hf(x)=f(x-h)$, with $f:\mathbb{R}^N\to\mathbb R$ in $L^\infty$ and $x,h\in\mathbb R^n$. Prove translations of $f$ by $h_n\to0$ converge weakly-* in $L^\infty$.

I manated it for functions a.e. equal to a.e. continuous functions. I was wondering:


*

*Is there any function that does NOT have the property of being a.e. equal to an a.e. continuous function?

*Does not having that property imply either non-measurability or not being a.e. bounded?


"a.e." means "Lebesgue-a.e.".
 A: The answer is NO to both questions.

Indeed, to give a counter-example, it is sufficient to show that there exists a bounded measurable function $f$ such that, for all negligible sets $N$, $f|_{\mathbb R^d\setminus N}$ is not continuous (in the subspace topology).

Let $\mathbb Q^d=\{q_n\}_{n\in\mathbb N}$ the $d$-uples of rational numbers. Consider $$B_n=B(q_n,2^{-n})=\left\{x\in\mathbb R^d\,:\,|x-q_n|<2^{-n}\right\}$$
Let $V:=\bigcup_{n\in\mathbb N} B_n$ and let $\chi_V$ its indicator function. Let $\mu$ the Lebesgue measure on $\mathbb R^d$ and let $\omega=\mu(B(0,1))$.
You can easily see that $\frac\omega2\le \mu(V)\le \sum_{n=1}^{\infty} 2^{-n}\omega=\omega$
So $\mu(\mathbb R^d\setminus V)=+\infty$

Since $\chi_V$ is identically $1$ on $V$ and identically $0$ on $\mathbb R^d\setminus V$, there does not exist a negligible set $N$ such that $\chi_V|_{\mathbb R^d\setminus N}:\mathbb R^d\setminus V\longrightarrow \{0,1\}$ is continuous.

Indeed, $R^d\setminus N$ must intersect both $V$ and $\mathbb R^d\setminus V$. Let $x\in (R^d\setminus V)\setminus N$.
There exists a sequence $q_{n_k}\in\mathbb Q^d$ such that $q_{n_k}\to x$ in $\mathbb R^d$. Notice that none of the $B_{n_k}$ can be containbed in $N$ (because $\mu(N)=0$). So there exists a sequence $a_k\in V\setminus N$ such that $$\begin{cases} a_k\in B_{n_k}\setminus N\subseteq V\setminus N\\
a_k\to x\end{cases}$$
But $\chi_V(x)=0$ and  $\chi_V(a_k)=1$. So $\chi_V|_{\mathbb R^d\setminus N}$ is not continuous.
Notice that $\chi_V$ is not only measurable, but also borelian. And it is bounded.
A: MickG, the answer to both of your questions is no (at least in the presence of measurability). Consider all bounded measurable functions which are supported on the hypercube $[0,1]^N$. If you assertions were true, $L_\infty[0,1]^N$ would be Banach-space isomorphic to $C[0,1]^N$ but the former space is non-separable. 
