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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:

  1. Is there any function that does NOT have the property of being a.e. equal to an a.e. continuous function?
  2. Does not having that property imply either non-measurability or not being a.e. bounded?

"a.e." means "Lebesgue-a.e.".

MickG
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  • There are plenty of functions that are continuous nowhere, and these will not be a.e. equal to an a.e. continuous function. For example $\chi_{\mathbb{Q}}$. – Paul Nov 11 '15 at 17:21
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    Wait. @Paul the indicator of the rationals is 0 a.e., since the rationals have measure 0, and 0 is certainly continuous. – MickG Nov 11 '15 at 17:22
  • this function is not continuous on the rationals, or irrationals. Or perhaps you are using a different definition of a.e. continuous than I am familiar with. – Paul Nov 11 '15 at 17:37
  • This is a bit of a wild guess but let $A$ be the set of real numbers that contains a $1$ in its decimal expansion and consider $\chi_A$. Upfront this is not continuous at any point outside $A$. Can you do any change a.e. to make it continuous? – Anguepa Nov 11 '15 at 17:38
  • @Paul I said a.e. equal to an a.e. continuous function. That function is a.e. equal to 0, which is everywhere continuous. Am I missing something? Evidently, a.e. equal to a.e. continuous does not imply a.e. continuous... – MickG Nov 11 '15 at 17:38

2 Answers2

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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.

  • I don't really see how you deduced the existence of $a_k$. – MickG Nov 11 '15 at 18:01
  • @MickG Actually, pick any sequence $a_k$ such that $\forall k,\ a_k\in B_{n_k}\setminus N$ (we know that there exists at least one because $\mu(N)$ cannot contain balls). By construction, $$|a_k-q_{n_k}|<2^{-n_k}\stackrel{k\to\infty}{\longrightarrow} 0$$ Since $q_{n_k}\to x$, then $a_k$ must tend to $x$ as well. –  Nov 11 '15 at 18:07
  • Imprecision in my previous commen: $N$ cannot contain balls, not $\mu(N)$ (I was writing $\mu(N)=0$, which is the reason why it it can't). –  Nov 11 '15 at 18:15
  • OK. So it is not a.e. continuous. But is it possible that, changing it in a null set, I get an a.e. continuous function, as happens with $\chi_{\mathbb Q}$ which is nowhere continuous but a.e. 0? – MickG Nov 11 '15 at 18:43
  • I get your definition of almost-everywhere continuous function is: a function $f$ is continuous almost everywhere if there exists a null set $N$ such that $f|_{\mathbb R^d\setminus N}$ is a continuous as a function $\mathbb R^d\setminus N\to \mathbb R$. A function almost-everywhere equal to an almost-everywhere continuous function is almost-everywhere contuinuous, because union of two null sets is null. –  Nov 11 '15 at 20:18
  • Perhaps I'm making some sort of confusion, but $\chi_{\mathbb{Q}}$ is definitely not a.e. continuous (it is nowhere continuous), yet it is a.e. equal to a constant which is everywhere continuous… didin't I just disprove your statement? – MickG Nov 11 '15 at 20:22
  • Give your definition of almost-everywhere continuous. –  Nov 11 '15 at 20:24
  • I think I've pinned it. I was thinking of "a.e. continuous" as "a.e. point is a point of continuity for $f:\mathbb{R}^n\to\mathbb{R}$", which is different from your definition. – MickG Nov 11 '15 at 20:24
  • Well, the definition I used of "almost-everywhere continuous" is weaker than yours (it includes your case and it is closed under almost-everywhere equivalence). So my counter-example should work regardless. –  Nov 11 '15 at 20:31
  • Indeed. Which is why I accepted. Too bad this raises another problem… :). – MickG Nov 11 '15 at 20:42
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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.

Tomasz Kania
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  • I do not follow those implications... I know $L^\infty$ is non-separable, but how would the presence of a function as in point 1 imply the two spaces would be B-s isomorphic? – MickG Nov 11 '15 at 17:44
  • If every bounded measurable function on $[0,1]^\infty$ were continuous a.e. then for each equivalence class $[f]\in L_\infty[0,1]^N$ we could choose such function, so the formal inlusion $C[0,1]^N\to L_\infty[0,1]^N$ would be onto. – Tomasz Kania Nov 12 '15 at 12:07