$L^{1}$ norm of a horizontally shifted measurable function

Question

Suppose we are in $(\mathbb{R}, \mathcal{B}(\mathbb{R}), m)$, where $m$ is Lebesgue measure and $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$.

Also, suppose $g: \mathbb{R} \to [-\infty, \infty]$ is in $L^{1}(dm)$.

I want to understand why $\int \limits_{\mathbb{R}} |g(y)| dy = \int \limits_{\mathbb{R}} |g(y-x)| dy$. That is, for each $x \in \mathbb{R}$, $||g(x)||_{1} = ||g(y - x)||_{1}$ holds.

Is this an obvious fact? I realize that $g(y - x)$ is measurable for each $x \in \mathbb{R}$.

Answer 1

For fixed $x \in \mathbb{R}$, define a mapping

$$\tau_x: \mathbb{R} \to \mathbb{R}, y \mapsto y-x.$$

Then $\tau_x$ is continuous (hence measurable) and $\tau_x^{-1} = \tau_{-x}$. By definition,

$$\int_{\mathbb{R}} |g(y-x)| \, m(dy) = \int_{\mathbb{R}} |g \circ \tau_x(y)| \, m(dy).$$

We can rewrite the right-hand side using image measures:

$$ \int_{\mathbb{R}} |g \circ \tau_x(y)| \, m(dy) = \int_{\mathbb{R}} |g(z)| (\tau_x m)(dz) \tag{1}$$

where

$$(\tau_x m)(B) := m(\tau_x^{-1}(B)) = m(\tau_{-x}(B)) = m(B+x)$$

denotes the image measure of $\tau_x$ with respect to the Lebesgue measure $m$. Since the Lebesgue measure is translation invariant, i.e.

$$m(B+x) = m(B)$$

for any Borel set $B \in \mathcal{B}(\mathbb{R})$, we conclude $\tau_x m = m$. Hence, by $(1)$,

$$ \int_{\mathbb{R}} |g(y-x)| \, m(dy) = \int_{\mathbb{R}} |g(z)| \, m(dz)$$