Does the equality $\lambda (\pi ^{-1}(B+g))=\lambda (\pi^{-1}(B))$ holds for any $B\in \mathfrak{B}_G$ and $g\in G$?

Let $\mathfrak{B}_G$ be the Borel $\sigma$-algebra of a topological abelian group $(G,+)$ and $\pi :\mathbb{R}^n\to G$ be a surjective measurable function.

Let $\lambda :\mathfrak{B}_n\to\overline{\mathbb{R}}$ be the Lebesgue measure of $\mathbb{R}^n$ in which $\mathfrak{B}_n$ is the standard Borel $\sigma$-algebra of $\mathbb{R}^n$.

My question is: Does the equality $\lambda (\pi ^{-1}(B+g))=\lambda (\pi^{-1}(B))$ holds for any $B\in \mathfrak{B}_G$ and $g\in G$?

I tried to prove that equality using the indicator function and the notion of pushforward measure:

Define $\mu :=\pi_\star \lambda $.

Let $B\in B_G\subseteq \pi_\star\mathfrak{B}_n$ and $g\in G$ be any elements. Then

$$\lambda (\pi ^{-1}(B+g))=\mu(B+g)=\int _{G}\mathbf{1}_{B+g}d\mu=\int _{\mathbb{R}^n}\mathbf{1}_{B+g}(\pi (x))d\lambda (x)\,\,\color{red}{(1)} $$

It's easy to see that $\mathbf{1}_{B+g}(\pi (x))=\mathbf{1}_{B}(\pi (x)-g)$.

If we can prove that there's $\alpha\in\mathbb{R}^n$ such that $\mathbf{1}_{B}(\pi (x)-g)=\mathbf{1}_{\pi^{-1}(B)+\alpha}(x)\,\,\color{red}{(2)}$ for all $x\in \mathbb{R}^n$, then we can use $(1)$ to prove that $\mu (B+g)=\lambda (\pi ^{-1}(B)+\alpha )=\lambda (\pi^{-1}(B))$. However I wasn't able to prove that there's such $\alpha\in\mathbb{R}^n$.

Thank you for your attention.

I'm asking the question above to prove that

As written, this is false. Take the case $n=1$ and the group $(\mathbb{Z},+)$ where we have given $\mathbb{Z}$ the discrete topology so that $\mathcal{B}(\mathbb{Z}) = 2^{\mathbb{Z}}$ and the measurable function to be $$ \pi(x) = \begin{cases} \lceil x\rceil \,\,\, x< 0\\ \lfloor x\rfloor \,\,\, x \geq 0 \end{cases} $$ where $\lceil x\rceil$ is the ceiling function and $\lfloor x\rfloor$ is the floor function. Then $\pi^{-1}(\{0\}) = (-1,1)$ while $\pi^{-1}(\{1\})=[1,2)$. Hence $$\lambda(\pi^{-1}(\{0\})) = 2\not= 1 = \lambda (\pi^{-1}(\{1\})) = \lambda(\pi^{-1}(\{0\} + 1))$$