Suppose $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)\to 0$ as $x\to-\infty$, $f(x)\to 1$ as $x\to \infty$, and $$g(t):=\int_{\mathbb{R}} [f(x+t)-f(x)]~dx$$ exists for all real $t$. Does it follow that $g(t)=t$ for all $t\in \mathbb{R}$?
Versions of this question have floated around the site a few times, and evidently is true under seemingly-mild assumptions on $f$: (1), (2). However, most impose constraints on $f(x)$ which are not necessary (even if they are sufficient):
Choosing $f$ as the Heaviside step function yields $g(t)=t$, so continuity is not necessary
Adding $e^{-x^2}$ to the previous example eliminates monotonicity, so $g(t)$ need not be a probability distribution
Alternatively, the answers don't make clear the assumptions being made. (Indeed, the accepted answer for question (1) makes clear that the level of rigor is unclear and asks for comment.) Since my foundations in this area are thin, I'm also being deliberately ambiguous about which definition of integral---e.g., improper Riemann vs Riemann-Stieltjes vs Lebesgue integrable---is being used in case it matters.