Let $f:\mathbb{R}^n\to \mathbb{R} $ is a Lebesgue measurable function and $T\in Gl(n,\mathbb{R})$ then show that $f\circ T$ is also Lebesgue measurable .
For Borel measurable functions it's easy to show . I think we need to use the fact that if $E$ is a Lebesgue measurable set then $E=B\cup N$ where $B$ is a $F_\sigma$ set i.e Borel set and $N$ is set with outer measure $0$ and $N\subset F$ where $F$ is a Borel set with measure zero , last line comes from the fact that Lebesgue measure is completion of Borel Measure .
There is similar question : Let $f:\mathbb{R}^n\to \mathbb{R} $ is a Lebesgue measurable function and $T: \mathbb{R}^n \to \mathbb{R}^n$ a $C^1$ diffeomorphism , then show that $f\circ T$ is also Lebesgue measurable .
Pick a Borel function $g$ on $\mathbb{R}^n$ such that $g = f$ ae. Let $N \subseteq \mathbb{R}^n$ be the null set where this equality fails. Then $M := T^{-1}(N)$ is a null set: If $\lambda$ is Lebesgue measure, then $$\lambda( T^{-1}(N) ) = | \det T^{-1} | \lambda(N) = 0$$ since $T$ is linear and invertible. For $x \notin M$ one has $Tx \notin N$, hence $g(Tx) = f(Tx)$. So $g \circ T = f \circ T$ ae. and $g \circ T$ is Borel by reasons you know. So $f \circ T$ is measurable since it equals ae. the Borel function $g \circ T$.