# Conditions on real function $f$ for $\int_{\mathbb{R}} [f(x+t)-f(x)]dx=t$ for all real $t$

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.

• Should the assumption on $f$ read $f(x)\rightarrow 1$ as $x\rightarrow\infty$? Commented Jul 31 at 2:47
• If $f$ is integrable at $-\infty$ and that integral exists for $t$, then $g(t) = \lim_{R\to\infty}\int_{-\infty}^R[f(x+t)-f(x)]\,dx = \lim_{R\to\infty}\int_{R}^{R+t}f(x)\,dx$. Now, if $f$ is continuous, then $g(t) = t\lim_{R\to\infty}f(\xi_R)$ by the mean value theorem for integrals. Here, $\xi_R\in (R,R+t)$. Hence, if you really mean that $f(x)\to\infty$ as $x\to\infty$, $g(t)$ cannot exist, which is a contradiction. Therefore, I think you mean $f(x)\to 1$ as $x\to\infty$, which then implies $g(t) = t$. Commented Jul 31 at 2:59
• @CW279 Blah, yes. Fixed. Commented Jul 31 at 2:59
• @amsmath Setting aside the wrong limit (which I just fixed) I agree that continuity is sufficient. But it's not necessary, since the unit step function would also yield $g(t)=t$. Commented Jul 31 at 3:00

Let's start with a proof that does not need any more assumptions than those already given, i.e. $$\lim_{x\to\infty} f(x) = A$$ and $$\lim_{x\to-\infty}f(x) = B$$. The case in the question corresponds to the special case $$A=1$$ and $$B=0$$. The integrals below are Riemann integrals.
We start with \begin{align}g_{R_1R_2}(t) &\equiv \int_{-R_2}^{R_1} [f(x+t) - f(x)]\,{\rm d}x\\ &= (A-B)t + \int_{R_1}^{R_1+t} [f(x)-A]\,{\rm d}x - \int_{-R_2}^{-R_2+t}[f(x)-B]\,{\rm d}x,\end{align} where we have taken full advantage of all the cancellation that happens between the two terms in the integrand. This shows that $$|g_{R_1R_2}(t)-(A-B)t| \leq |t|(\sup_{x\in[R_1,R_1+t]}|f(x)-A| + \sup_{x\in[-R_2,-R_2+t]}|f(x)-B|).$$ The existence of the limits at $$\pm\infty$$ ensures that the supremum above has to be finite for large enough $$R_1,R_2$$ and goes to zero as $$R_1,R_2\to\infty$$ which gives us the desired result $$g(t) \equiv \lim_{R_1,R_2\to\infty}g_{R_1R_2}(t) = (A-B)t.$$
We see that it is only the behavior of $$f$$ close to $$\pm\infty$$ that matters (i.e. that the limits exists) as long as the integral $$g(t)$$ actually exists. The integral will for example exist for all functions $$f$$ that are integrable on compact subsets of $$\mathbb{R}$$ which is a pretty weak condition.