A measurable implies $f^{-1}[A]$ measurable

Here is a counterxample with the Lebesgue measure on the real line.

Let $f:[0,1] \to \Bbb{R}$ be the Cantor-Lebesgue function,AKA the Devil's staircase.

Define $g:[0,1] \to [0,2]$ such that $g(x)=f(x)+x$.

It is not difficult to see that $g$ is a homeomorphism.

And also $g(Cantor)$ has a positive Lebesgue measure thus exists $V \subseteq g(Cantor)$ which is non-measurable.

So $g^{-1}(V) \subseteq Cantor$ thus is measurable as a set of measure zero and also $g^{-1}$ is continuous.

But $(g^{-1})^{-1}[g^{-1}(V)]=V$ which is non-measurable.

Without more context, of course not. If we consider the identity map $\newcommand{\id}{\text{id}}\id : Y\to Y$ (with $Y$ having a nontrivial topology) with the first $Y$ having the trivial $\sigma$-algebra and the second $Y$ having the Borel $\sigma$-algebra, then if $\newcommand{\cB}{\mathcal{B}}\newcommand{\nullset}{\varnothing}\nullset\subsetneq U\subsetneq Y$ is an open subset of $Y$, then $\newcommand{\inv}{^{-1}}\id\inv(U)=U$ is not measurable in the first copy of $Y$, even though it is measurable in the second copy of $Y$.

If both spaces have the Borel $\sigma$-algebra, then the answer is yes.

Proof: First we need a fact about $\sigma$-algebras generated by a collection of sets, since the Borel $\sigma$-algebra is the $\sigma$-algebra generated by the open sets.

Lemma: Let $f:X\to Y$ be a map of sets. If $S\subseteq\mathcal{P}(Y)$ generates a $\sigma$-algebra on $Y$, $\cB$, then the preimage of $S$ generates the preimage of $\cB$.

Proof of lemma: Since preimages preserve intersections, unions, complements, the null set and the whole set, for any function of sets from $f:X\to Y$, and any $\sigma$-algebra $\cB$ on $Y$, the set $\newcommand{\set}[1]{\left\{{#1}\right\}}f\inv\cB:=\set{f\inv(A) : A\in\cB}$ is a $\sigma$-algebra on $X$ called the preimage of $\cB$. If $S$ generates $\cB$, then certainly $f\inv S \subseteq f\inv \cB$, so we just need to show that if $\newcommand{\cA}{\mathcal{A}}f\inv S \subseteq \cA$ for $\cA$ a $\sigma$-algebra on $X$, then $f\inv\cB\subseteq\cA$. For this, consider $f_*\cA :=\set{ B : f\inv(B) \in \cA}$ (you can verify that this is a $\sigma$-algebra on $Y$). In particular, note that $S\subseteq f_*\cA$, since $f\inv S\subseteq \cA$, so since $\cB$ was generated by $S$, $\cB\subseteq f_*\cA$, which means that $f\inv\cB\subseteq \cA$. Hence $f\inv\cB$ is a subset of all $\sigma$-algebras containing $f\inv S$. Thus it is the $\sigma$-algebra generated by $f\inv S$.

Resuming main proof: Now we apply the lemma to the situation of a continuous map between topological spaces with their Borel $\sigma$-algebras. Let $f: X\to Y$ be continuous, let $\cB(X)$ and $\cB(Y)$ be the Borel $\sigma$-algebras on $X$ and $Y$ respectively, and let $\tau_X$ and $\tau_Y$ be the topologies on $X$ and $Y$ respectively. Then by the lemma above, $f\inv \cB(Y)$ is generated by $f\inv\tau_Y$. Since $f$ is continuous, $f\inv\tau_Y\subseteq \tau_X\subseteq\cB(X)$, so $f\inv\cB(Y)\subseteq \cB(X)$. Unravelling the definitions, this is exactly what it means for $f$ to be measurable when $X$ and $Y$ are equipped with their respective Borel $\sigma$-algebras.

Point of caution! The key point to be careful of here is that the Lebesgue measure is strictly larger than the Borel measure on $\newcommand{\RR}{\mathbb{R}}\RR^n$. Hence if $f:\RR^m\to \RR^n$ is continuous, and $\cB^n$ and $\newcommand{\cL}{\mathcal{L}}\cL^n$ are the Borel and Lebesgue $\sigma$-algebras on $\RR^n$ respectively, while $f:(\RR^n,\cB^n)\to(\RR^m,\cB^m)$, and $f:(\RR^n,\cL^n)\to(\RR^m,\cB^m)$ are measurable, $f:(\RR^n,\cL^n)\to(\RR^m,\cL^m)$ and $f:(\RR^n,\cB^n)\to(\RR^m,\cL^m)$ may not be. In his answer, Marios Gretsas has provided an example of such an $f$ which is continuous but not Lebesgue-Lebesgue measurable.