I'm going to present the following problem and a solution given in an older colloquium in Measure Theory in hope of understanding a statement in the final conclusion.
$\boldsymbol{\underline{\text{Question:}}}$
Suppose $f:\Bbb R\to\Bbb R$ is a continuous function and let $\mathcal L$ denote the Lebesgue $\sigma$ - algebra on $\Bbb R.$ Prove or disapprove the following assertion: $$f(\mathcal L):=\left\{f(A)\mid A\in\mathcal L\right\}\subseteq\mathcal L.$$
$\boldsymbol{\underline{\text{Answer.}}}$ The statement doesn't hold. Indeed, define $f:\Bbb R\to\Bbb R$ by $$f(x)=\begin{cases}0,&\text{ if } x<0\\ c(x), &\text{ if }x\in [0,1]\\ 1,&\text{ if } x>1,\end{cases}$$
where $c:[0,1]\to [0,1]$ is Cantor's function. Then, $f$ is continuous and $f(C)=[0,1]$ for Cantor's set $C.$ Let $B\subseteq (0,1)$ s. t. $B\notin\mathcal L.$ Let's consider the set $A:=C\cap f^{-1}(B).$ Since $A\subseteq C,$ and, from the lectures, we know that $\lambda(C)=0,$ and since $(\Bbb R,\mathcal L,\lambda)$ is complete, $A\in\mathcal L$. Since $f(C)=[0,1],$ we conclude that $f(A)=B$.
I have a problem understanding why $f(C)=[0,1]$ and $A=C\cap f^{-1}(B)$ implies $f(A)=B,$ since, in general $f(X\cap Y)\subseteq f(X)\cap f(Y),$ so $f(A)\subseteq f(B)$. According to what I've looked up, the Cantor function doesn't seem to be injective, only continuous and surjective, as shown in threads $(1)$,$(2)$, $(3)$.
I initially thought two seemingly different definitions (laid here and here) of the so-called Cantor function might be different and be the cause of my misunderstanding, but they turn out to be equivalent.
My question is, is non - injectivity a problem in this particular argument?
Just in case, I'll mention I'm aware of a (continuous) Borel - measurable function $f:[0,1]\to[1,2]$ that we constructed in our lectures in a similar way as it is in Donald L. Cohn's Measure Theory:
with the exception that we defined $f:[0,1]\to[1,2], f(x):=c(x)+x$ and took $A:=f^{-1}(S),$ where $S$ is a non - Lebesgue measurable subset of $f(C),$ where $C\subseteq [0,1]$ is the Cantor set and $A\notin\mathcal B(\Bbb R).$
"I have a problem understanding why $f(C)=[0,1]$ and $A=C\cap f^{-1}(B)$ implies $f(A)=B$": $B\subset f(C)\implies f(\{x\in C\mid f(x)\in B\})=B.$