# If $\int_{-\infty}^{+\infty} |f(t)|dt < \infty$, show that $\int_{-\infty}^{+\infty}f(t)dt$ exists

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.

Attempt:

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?

• Seems fine @Andromeda. Sep 22, 2022 at 10:27
• 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. Sep 22, 2022 at 10:47
• Doesn't this just follow instantly from the triangle inequality for integrals , $$\int_E |f|\mathrm d\mu\leq\left|\int_E f\mathrm d\mu\right|$$ Sep 22, 2022 at 19:03
• @K.defaoite How then? Sep 22, 2022 at 19:56