Does absolute convergence of an integal imply that it is equal to the limit of the integral on $[-c,c]$ as $c\to \infty$?

Question

I am asking whether: $$\int_\mathbb{R} |f(x)| dx<\infty \implies \int_\mathbb{R} f(x) dx= \lim_{c\to \infty} \int_{-c}^c f(x) dx$$.

I think that it's true, and it feels like the continuous analogue of the Riemann rearrangement theorem, in the sense that it says the way you add up the terms (more appropriately here, integrate) does not matter, you will still get the same result.

Let's assume that the integrals are of the Riemann variety, but I have a feeling that it holds for Lebesgue integrals as well.

The functions I am working with are continuous and do not have any "problematic" points in $\mathbb{R}$.

Answer 1

It follows from dominated convergence theorem. Write the truncated integral as : $$I_c = \int_{\mathbb{R}} f(x) \mathbb{1}_{[-c, c]}(x) dx$$

Define $g(x, c) = f(x) \mathbb{1}_{[-c, c]}(x)$. Then :

1. $\forall x \in \mathbb{R}, ~ g(x, c) \underset{c \to \infty}{\to} f(x)$
2. $\forall (x, c) \in \mathbb{R}^2, ~ |g(x, c)| \leq f(x)$. And $\int_{\mathbb{R}} |f(x)| dx < \infty$.

Thus, we can apply the dominated convergence theorem that states that $I_c \underset{c \to \infty}{\to} \int_{\mathbb{R}} f(x) dx$