Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is differentiable and both $f(x)$ and $f^{\prime}(x)$ are integrable on $\mathbb{R}$. Show $\int_{\mathbb{R}} f^{\prime}(x) \, dx = 0$.
I know that if $f^+$ and $f^-$ are the positive and negative parts of $f$, then
$$\int f = \int f^+ - \int f^-$$
Also, $|f| = f^+ + f^-$ so we know that $f$ is integrable iff $\int |f| < \infty$.
We can also treat the improper integral like so:
$$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{a} f(x) \, dx + \int_{a}^{+\infty} f(x) \, dx$$
provided both integrals on the right converge. I'm trying to take this information and formulate a proof out of it but it just won't come together. Any help would be appreciated.
Since $f$ is differentiable and $f'$ is integrable, by the fundamental theorem of calculus we have
$$ \int_{\mathbb{R}} f'(x) \mathbf{1}_{[a,b]}(x) \, dx = \int_{a}^{b} f'(x) \, dx = f(b) - f(a) $$
By the dominated convergence theorem, as $a\to-\infty$ and $b\to\infty$ the above quantity converges, from which we know that both
$$ \lim_{a\to-\infty} f(a) \qquad \text{and} \qquad \lim_{b\to\infty} f(b) $$
converge. Now by the integrability of $f$ it is easy to conclude that both must be zero. Therefore
$$ \int_{\mathbb{R}} f'(x) \, dx = \left( \lim_{b\to\infty} f(b) \right) - \left( \lim_{a\to-\infty} f(a) \right) = 0. $$