Given $\lim_{x \rightarrow \infty} f(x) = 2$ and $\lim_{x \rightarrow -\infty} f(x) = 1$, what's the following equal to: $\large \int\limits_{-\infty}^\infty (f(x+1)-f(x)) dx $
I know that $\large \int\limits_{-\infty}^\infty dx $ is exists if and only if both $\large \int\limits_{0}^\infty dx $ and $\large \int\limits_{-\infty}^0 dx $ exist.Then I tried splitting the integrand into two but I realized that I can't do it since the limit of the given function isn't zero.
Any hints?
First, if one were to be really pedantic, there exists a bounded Lebesgue-measurable function $f:\mathbb{R}\to\mathbb{R}$ such that $\lim_{x \rightarrow \infty} f(x) = 2$ and $\lim_{x \rightarrow -\infty} f(x) = 1$ but such that $g(x)=f(x+1)-f(x)$ is not Lebesgue-integrable over $\mathbb{R}$.
Let's instead suppose that $f$ is Riemann-integrable over every bounded real interval, and we are trying to show that if $N_i$ and $M_j$ are sequences that tend to $\infty$, then the double sequence $\displaystyle\int_{-M_j} ^{N_i} (f(x+1)-f(x))\, \text{d}x$ converges to a limit. If you don't know what it means for a double sequence to converge to a limit, see the accepted answer in
What is the definition of double sequence $a_{mn}$ being convergent to $l$?
Now, for $N_i$ and $M_j$ both positive, we have
$\displaystyle\int_{-M_j} ^{N_i} (f(x+1)-f(x))\, \text{d}x$
$=\displaystyle\int_{-M_j} ^{N_i} (f(x+1))\, \text{d}x-\int_{-M_j} ^{N_i} f(x)\, \text{d}x$
$=\displaystyle\int_{-M_j+1} ^{N_i+1} f(x)\, \text{d}x-\int_{-M_j} ^{N_i} f(x)\, \text{d}x$
$=\displaystyle\int_{N_i} ^{N_i+1} f(x)\, \text{d}x-\int_{-M_j} ^{-M_j+1} f(x)\, \text{d}x$
which of course tends to $2-1=1$.
