I‘m asked to find an example of a $\mathcal{L}^1$-measurable function $f : [0,1] → \mathbb{R}$ such that for every $\mathcal{L}^1$-measurable set $M ⊂ [0,1]$ with $\mathcal{L}^1(M) = 1$, the restriction $f|_M : M → \mathbb{R} $ is discontinuous in all but finitely many points of $M$.
I know that I should use the existence of a Lebesgue measurable subset $A ⊂ [0, 1]$ such that $\mathcal{L}^1(U ∩ A) · \mathcal{L}^1(U ∩ A^C) > 0$ for all nonempty open subsets $U ⊂ [0, 1]$.
In the provided solution they state that if we chose $f$ the characteristic function on this $A$, then $f|_M$ is always discontinuous except at $0$ and $1$. The proof I found is very different from the one in the solution manual and since I also proved discontinuity even at $0$ and $1$ I‘m a bit concerned wether I‘ve made a mistake. I‘d be really gratful if someone could check it:
Let $f$ be the characteristic function on $A$ and let $f|_M$ be continuous at $x \in [0,1]$, let‘s suppose $f(x)=1$. Then there exists an open neighbourhood $U$ of $x$ in $[0,1]$ s.t. $f(U \cap M) \subset (0.9,1.1)$ thus we have $U \cap M \subseteq A$ (because $f$ only takes the values $0,1$). Now because of $A$‘s properties: $$ 0< \mathcal{L}^1(U \cap A^C)= \mathcal{L}^1((U \cap A^C)\setminus M)+ \mathcal{L}^1(U \cap A^C \cap M) \leq \mathcal{L}^1(U\setminus M) $$ But we also have: $$ \mathcal{L}^1(U \setminus M) \leq \mathcal{L}^1([0,1]\setminus M)=1-1=0 $$
Which is a contradiction. The case $f(x)=0$ works exactly the same by switching the roles of $A$ and $A^C$.