Let $f: \mathbb{R}\to \mathbb{R}$ be a function such that $f$ and $|f|$ are Riemann-integrable on each closed interval. If $\int_{-\infty}^{+\infty} |f(t)|dt < \infty$, show that $\int_{-\infty}^{+\infty}f(t)dt$ exists.


We need to show that $\int_0^{+\infty} f(t)dt:= \lim_{x\to \infty}\int_0^x f(t)dt$ exists.

Let $\{x_n\}_n$ be a non-decreasing sequence with $x_n\nearrow \infty$. Then for $n \le m $, we have $$\left|\int_0^{x_m}f(t)dt-\int_0^{x_n} f(t)dt\right|\le \int_{x_n}^{x_m}|f(t)|dt = \int^{x_m}_0 |f(t)|dt - \int_0^{x_n}|f(t)|dt \stackrel{m,n \to \infty}\longrightarrow 0$$ so $\lim_{n \to \infty }\int_0^{x_n} f(t) dt$ exists.

Now, we use the following fact:

Let $g: \mathbb{R}\to \mathbb{R}$ a function. If for all $\{x_n\}$ non-decreasing sequences with $x_n \nearrow +\infty$, we have that $\lim_n g(x_n)$ exists, then $\lim_{x\to \infty} g(x)$ exists.

Similarly, we prove that $\int_{-\infty}^0 f(t)dt$ exists. Hence, $$\int_{-\infty}^{+\infty}f(t)dt$$ exists. Is my attempt correct?

  • 3
    $\begingroup$ Seems fine @Andromeda. $\endgroup$ Sep 22, 2022 at 10:27
  • 1
    $\begingroup$ Alternatively, $(|f|+f)/2$ and $(|f|-f)/2$ are nonnegative and the fact that the integral of $|f|$ exists, it follows that these nonnegative functions also have integrals. Their difference, $f$, must therefore be integrable. $\endgroup$ Sep 22, 2022 at 10:47
  • $\begingroup$ Doesn't this just follow instantly from the triangle inequality for integrals , $$\int_E |f|\mathrm d\mu\leq\left|\int_E f\mathrm d\mu\right|$$ $\endgroup$
    – K.defaoite
    Sep 22, 2022 at 19:03
  • $\begingroup$ @K.defaoite How then? $\endgroup$
    – Andromeda
    Sep 22, 2022 at 19:56


You must log in to answer this question.