$f(x) \in \mathcal{M} (\mathbb{R}, \mathfrak{B}) \Leftrightarrow g_y(x):=f(x+y) \in \mathcal{M} (\mathbb{R}, \mathfrak{B})$

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I want to show that

$$f(x) \in \mathcal{M} (\Bbb{R}, \mathfrak{B}) \Leftrightarrow g_y(x):=f(x+y) \in \mathcal{M} (\Bbb{R}, \mathfrak{B}), \ \ \forall y \in \Bbb{R}$$

'$\in \mathcal{M}$' means measurable

and that then

$$\int f \mathrm{d}\lambda = \int g_y \mathrm{d} \lambda$$ if both sides exist.

For the first part: Let $[f<c] := \{x|f(x)<c\}$

$$x \in [g_y<c] \Leftrightarrow = g_y(x) <c \Leftrightarrow f(x+y)<c \Leftrightarrow (x+y) \in [f<c] \Leftrightarrow x \in [f<c]-y$$

Therefore $$[g_y<c]=[f<c]-y$$

Now, because the borel-sigma-algebra is translation-invariant

$$[f<c] \Rightarrow [f<c]-y \Rightarrow [g_y<c]$$ is measurable.

Is that correct? And how do I show $\int f \mathrm{d}\lambda = \int g_y \mathrm{d} \lambda \ $ if both sides exist?