A Lebesgue measurable function $f:[0,1]\to R$ is always continuous on some set $ B \subseteq [0,1] $ and $\mu (B) = 1$ . True or false?

Question

A function $ f(x) $ is continuous on $ B$ if for $ a, a+h\in B $, $|f (a+h)-f (a)| \to 0$ as $|h| \to 0$ .

A Lebesgue measurable function $f:[0,1]\to R$ there is some set $ B \subseteq [0,1] $ and $\mu (B) = 1$ .There is a continuous function $ g: B \to R $ and $\mu ${$x \in B:f (x) \ne g (x)$} $=0$ . True or false ?

This looks like Lusin's theorem but is not exactly the same because Lusin's theorem does not guarantee $ \mu (B) = 1$

I believe it is false because if it were true :

One can easily show there is a sequence of schwartz functions {$\phi_n $} such that for $ p \ge 1$ and every measurable function $ f \in L^p$ , $\lim_{n \to \infty} f*\phi_n= f (x) $ almost everywhere

Hmm, sort of thinking along the lines of the indicator function of a fat Cantor set... Not sure if that actually works... — Daniel Schepler, Jun 18 '18 at 22:57
@DanielSchepler I think that might work but the problem is one has to use indicator function and end up to constant — ibnAbu, Jun 18 '18 at 23:09
The closest i could find http://mathworld.wolfram.com/LusinsTheorem.html — ibnAbu, Jun 18 '18 at 23:19

charmd · Accepted Answer · 2018-06-19T05:42:05.940

Claim: there exists a measurable set $A \subset [0,1]$ such that for all $0\le a<b\le 1$, $0 < \mu(A\cap]a,b[)<b-a$.

Now $f = \mathbf{1}_A$ is a solution. Assume ad absurdum that you can find $B$ such that $f_B$ is continuous and $\mu(B)=1$. The key point is that $A\cap B$ and $([0,1]-A)\cap B$ share at least one accumulation point. This is because of :

Lemma: every point in $B$ is an accumulation point of $A\cap B$ and of $([0,1]-A)\cap B$.

Proof: take $b\in B$. For every $\varepsilon > 0$, $\mu(A\cap B\cap ]b-\varepsilon,b+\varepsilon[)=\mu(A\cap ]b-\varepsilon,b+\varepsilon[)>0$ using the definition of $A$. Thus for all $\varepsilon>0$, $]b-\varepsilon,b+\varepsilon[\cap A\cap B$ is not empty. Using the second inequality in the definition of $A$, we get the same result in its complement.

We can conclude regarding the original problem. Take any $b\in B$. The previous lemma enables us to consider two sequences respectively in $f^{-1}(\{1\})\cap B$ and $f^{-1}(\{0\})\cap B$, converging towards $b$. Hence $f_B$ is not continuous at $b$, which is absurd.

thanks for the answer and the link. – ibnAbu Jun 19 '18 at 16:12 — ibnAbu, Jun 19 '18 at 16:12

score 0 · Answer 2 · answered Jun 18 '18 at 22:30

0

The Dirichlet function, which is 1 on a rational and 0 otherwise, is continuous nowhere.

answered Jun 18 '18 at 22:30

ncmathsadist

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2

$f_{|[0,1]-\mathbb{Q}}$ is continuous (because it is constant) – charmd Jun 18 '18 at 22:36

A Lebesgue measurable function $f:[0,1]\to R$ is always continuous on some set $ B \subseteq [0,1] $ and $\mu (B) = 1$ . True or false?

2 Answers2