Translation invariance of function

Question

For the measurable function $f\mathpunct{:}\mathbb{R}\to\mathbb{R}$ if it meets the requirement that for any real number $a$, the function $f (x+a) - f (x)=0 $ a.e. , does this necessarily imply that $f (x)= \text {constant} $ a.e. ?

If we assume that $f (x)$ is continuous at a certain point, let $$ N_q=\{ x\in \mathbb{R}\ |\ f(x+q)\neq f(x) \} $$ We can get that the measure of $N_ q$ is $0$. Let $$ B=\{ x\in \mathbb{R}\ |\ f(x+q)=f(x),\forall q\in \mathbb{Q} \} $$ clearly $$ B\subset \mathbb{R}-\bigcup_{q\in \mathbb{Q}} N_q $$ By continuity, we can obtain a proof that $f$ is a constant on $B$. However, how can we prove the general case?

Answer 1

$f^{+}$ and $f^{-}$ satisfy the same hypothesis, so we may suppose $f \geq 0$. We can then look at $e^{-f}$ to make $f$ bounded also.

If $\mu(E)=\int_E fd\lambda$ ($\lambda$ being the Lebesgue measure) then you can check that $\mu$ is transaltion invariant and finite on compact sets. This implies that $\mu =c \lambda$ for some constant $c$ and this gives $f=c$ a.e..

Answer 2

Take $F(x,a)=|f(x+a)-f(x)|$. By Tonelli $$\iint F(x,a)dadx=\iint F(x,a)dxda=0$$ So there exist some $x$ for which $F(x,a)=|f(x+a)-f(x)|$ for $a$ a.e.. This implies $f(y)=f(x)$ $y$ a.e. as $$\{y : f(x)=f(x)\}=\{a: f(a+x)=f(x)\} + x $$