A translation-invariant L1 function on the line is zero

First record the representative formula for $f\in L^1(\mathbb{R})$ and $h_0\in C_c(\mathbb{R})$: the function $x\mapsto F(x):={\int }_{\mathbb{R}}f(y)h_0(x-y)d{\lambda }_1(y)$ represents the class $f\ast h_0$, because for $u_n\in C_c$ with $\mathrm{\parallel }{u}_n-f{\mathrm{\parallel }}_1\to 0$ one has $u_n\ast h_0\to f\ast h_0$ by [F3], while $\mathrm{\mid }F(x)-(u_n\ast h_0)(x)\mathrm{\mid }\le {\int }_{\mathbb{R}}\mathrm{\mid }f-u_n\mathrm{\mid }(y)\mathrm{\mid }{h}_0(x-y)\mathrm{\mid }d{\lambda }_1(y)$ and hence $\mathrm{\parallel }F-u_n\ast h_0{\mathrm{\parallel }}_1\le \mathrm{\parallel }f-u_n{\mathrm{\parallel }}_1\mathrm{\parallel }{h}_0{\mathrm{\parallel }}_1\to 0$ by [F4] applied to the nonnegative product-measurable function $\mathrm{\mid }f-u_n\mathrm{\mid }(y)\mathrm{\mid }{h}_0(x-y)\mathrm{\mid }$ and translation invariance [F1], so $F$ and $f\ast h_0$ differ only on a null set.

Let $U_n:=(-1\mathrm{/}(n+1),1\mathrm{/}(n+1))$ and choose $e_n:=e_{U_n}$ from the net of [F2]; for every $n$ and every $x$ the representative of $g\ast e_n$ from step 1.1 gives $(g\ast e_n)(x)={\int }_{\mathbb{R}}g(y)e_n(x-y)d{\lambda }_1(y)={\int }_{\mathbb{R}}g(x-z)e_n(z)d{\lambda }_1(z)$ by the substitution $z=x-y$ and translation invariance, while the hypothesis with shift $t=x$ gives $g(x-z)=g(-z)$ for almost every $z$, so $(g\ast e_n)(x)={\int }_{\mathbb{R}}g(-z)e_n(z)d{\lambda }_1(z)=:c_n$ is a constant independent of $x$; thus $g\ast e_n$ equals the constant $c_n$ almost everywhere.

By [F3] each $g\ast e_n$ is an $L^1$ class. Step 2.1 identifies it with the constant $c_n$; since ${\lambda }_1(\mathbb{R})=\mathrm{\infty }$, integrability forces $c_n=0$. The neighbourhoods $U_n=(-1\mathrm{/}(n+1),1\mathrm{/}(n+1))$ are cofinal in the identity neighbourhoods: every such neighbourhood contains some $U_m$, and $U_n\subseteq U_m$ for $n\ge m$. The right approximate-identity convergence in [F2] therefore gives $\mathrm{\parallel }g\ast e_n-g{\mathrm{\parallel }}_1\to 0$. Since $g\ast e_n=0$ for every $n$, $\mathrm{\parallel }g{\mathrm{\parallel }}_1=0$ and $g=0$ almost everywhere.

Finally let $v\in L^2(\mathbb{R},{\lambda }_1)$ satisfy $\mathrm{\mid }v(x-t)\mathrm{\mid }=\mathrm{\mid }v(x)\mathrm{\mid }$ for every $t$ and almost every $x$; then $g:=\mathrm{\mid }v{\mathrm{\mid }}^2$ lies in $L^1(\mathbb{R},{\lambda }_1)$ by [F6] and satisfies $g(x-t)=g(x)$ for every $t$ and almost every $x$, so step 3.1 applied to this $g$ gives $g=0$ almost everywhere, that is, $v=0$ almost everywhere, which is the stated consequence; this proves the lemma. The Axiom of Choice is consumed through the approximate identity net of [F2] and through those measure-theoretic suppliers of [F1] that need it, the complete-measure, dilation-reflection and Radon-measure theorems being proved under the Axiom of Countable Choice; the translation, convolution and subsequence arguments are choice-free apart from those inputs.