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