$f :\mathbb R \to \mathbb R$ be a bijective Lebesgue measurable function , then is $f^{-1}:\mathbb R \to \mathbb R$ Lebesgue measurable?

Let $f :\mathbb R \to \mathbb R$ be a bijective Lebesgue measurable function , then is $f^{-1}:\mathbb R \to \mathbb R$ Lebesgue measurable ? I don't think this is true but I can't find any counterexample , or any way through . Please help . Thanks in advance

NOTE : Any counter example using non Lebesgue measurable set is fine with me

The interval $(0,1/4]$ contains a nonmeasurable subset (meaning a non Lebesgue measurable subset). The union of this set with $(1/4,1/2]$ is then a nonmeasurable subset $E$ of $(0,1/2]$ that has cardinality $c.$

Let $K_1,K_2$ be two disjoint Cantor sets, each a subset of $(0,1),$ having measure $0.$ Each of these has cardinality $c.$ Note that $(0,1)\setminus E$ also has cardinality $c.$ Thus there exists a bijection $f: K_1\cup K_2 \to (0,1)$ with $f(K_1) = E,$ $f(K_2)=(0,1)\setminus E.$

We can extend $f$ now to $K_1\cup K_2 \cup\mathbb Z$ by defining $f(n) = n$ for each $n\in \mathbb Z.$ Note that so far, we have only defined $f$ on a set (a Borel set in fact) of measure $0.$ This of course cannot harm Lebesgue measurability in any way as we move towards defining $f$ on all of $\mathbb R$ (although it could muck up Borel measurability).

Onwards: $K_1\cup K_2 \cup \mathbb Z$ is a closed subset of $\mathbb R.$ Hence its complement is the pairwise disjoint union of countably many open intervals $I_1,I_2, \dots.$ The idea is to map each of $I_1,I_3,I_5, \dots$ bijectively onto $(-1,0),(-2,-1),(-3,-2),\dots$ respectively. This can be done with homeomorphisms. Same with $I_2,I_4,I_6, \dots$ and the intervals $(1,2),(2,3),(3,4),\dots.$ Choose whatever homeomorphisms you like for this.

We now have $f$ defined as a bijection of $\mathbb R$ to $\mathbb R.$ And it is Lebesgue measurable, as you should verify.

Is $f^{-1}$ Lebesgue measurable? If it were, then $(f^{-1})^{-1}= f$ would take Borel sets to Lebesgue measurable sets. But as we've seen, $f(K_1) = E,$ and $E$ is not Lebesgue measurable.