The following problem is from my exercise sheet. I was able to write a solution, but I am unsure in some passages, mainly in the sup equations in the second part, and I would be very grateful if someone could verify my solution. Any comments and suggestions will be the most appreciated, likewise critics concerning the clarity and writing style.
Prove that, if $X \in \mathcal{M}(\mathbb{R}^N)$ such that $m(X) < \infty, \ y\in \mathbb{R}^N \text{ and } f:y+X \longrightarrow \mathbb{R}$ is a $\mathcal{M}(y+X)$-measurable function, then the map $g:X \longrightarrow \mathbb{R}, x \mapsto f(x+y)$ is measurable and $$ \int_Xf(x+y)dm(x) = \int_{X+y}f(x)dm(x) $$ where $y+X \doteq \{y+x ; \ x \in X \}$, $\mathcal{M}$ is the Lebesgue $\sigma$-algebra and m is the Lebesgue measure on $\mathbb{R}^N$.
My solution is as follows:
Since f is $\mathcal{M}(y+X)$-measurable, then, given any $\alpha \in \mathbb{R}$: $$ \{ x+y \in \mathbb{R}^N; \ f(x+y) < \alpha \} \in \mathcal{M}(y+X) = \{ (y+X) \cap A; \ A \in \mathcal{M}(\mathbb{R}^N) \} $$ $$ \therefore \{ x+y \in \mathbb{R}^N; \ f(x+y) < \alpha \} \in \mathcal{M}(\mathbb{R}^N) $$ $$ \therefore \{ x \in \mathbb{R}^N; \ f(x+y) < \alpha\} = -y + \{ x+y \in \mathbb{R}^N; \ f(x+y) < \alpha \} \in \mathcal{M}(\mathbb{R}^N), $$ since translated Lebegue measurable sets remain measurable. But then $$ \{ x \in \mathbb{R}^N; \ f(x+y) < \alpha\} \in \{X\cap A;\ A \in \mathcal{M}(\mathbb{R})^N \} = \mathcal{M}(X) $$ $$ \therefore \{ x \in \mathbb{R}^N; \ f(x+y) < \alpha\} = \{ x \in \mathbb{R}^N; \ g(x) < \alpha\} \in \mathcal{M}(X) $$ Since $\alpha$ in any real number, we have that g is $\mathcal{M}(X)$-measurable.
Now, since $g$ is measurable and limited, there exists the number$\int_X g(x)dm(x)$ and, by definition of the integral: \begin{equation*} \int_Xf(x+y)dm(x) \! \begin{aligned}[t] & = \int_X g(x)d\text{m}(x) \\ & = \sup\left\{\int_X \phi(x)dm(x); \phi \text{ is simple }, \phi(x) \leq g(x) \forall x \in X \right\} \\ & = \sup\left\{\int_X \Phi(x+y)dm(x); \Phi \text{ is simple }, \Phi(x+y) \leq f(x+y) \forall x \in X \right\} \\ & = \sup \left\{\int_{y+X} \psi(t)dm(t); \psi \text{ is simple }, \psi(t)\leq f(t) \forall t \in y+X \right\} \\ & = \int_{y+X}f(x) dm(x) \end{aligned} \end{equation*}