Continuity of a Lebesgue indefinite integral over unbounded interval

Question

We know that if $f : [a,b] \rightarrow \mathbb{R}$ is Lebesgue-integrable, then

$$ F(x) = \int_{a}^{x} f(t) dt $$

is continuous. But if $f : \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue-integrable, it is true that

$$ F(x) = \int_{-\infty}^{x} f(t) dt $$

is continuous?

Note that now $f$ is defined on unbounded interval, and I don't have any idea to start to think. Any hint will be hellpful.

Answer 1

Let $a$ and $b$ be real numbers such that $a < b$. Then for $x \in [a,b]$,

$$F(x) = \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^{a} f(t) dt + \int_{a}^{x} f(t) dt$$

is a constant plus a continuous function. $F$ is therefore continuous on $[a,b]$. And a function that is continuous on every finite interval is continuous on $\mathbb R$.

Answer 2

Here is how.

$$ |F(x+h) - F(x)|= \bigg|\int_{-\infty}^{x+h} f(t) dt -\int_{-\infty}^{x} f(t) dt \bigg| \leq \int_{x}^{x+h} |f(t)| dt < \epsilon $$