# Absolutely integrable function on $\Bbb R$

Let $$f:\Bbb R \to \Bbb R$$ be a function such that $$\int_{-\infty}^\infty \vert f(x) \vert < \infty$$ and let $$F:\Bbb R \to \Bbb R$$ be such that $$F(x)=\int_{-\infty}^x f(t)dt$$. Then which of the following is/are true?

• $$f$$ is continuous

• $$F$$ is uniformly continuous

• $$f$$ is bounded

My Try

$$f$$ need not be bounded, for example, consider $$f(x)=e^{-x^2}$$. Also $$f$$ need not be continuous , for example , take $$f(x)=\begin{cases} 0 &\text{if}\;-\infty

$$F$$ is uniformly continuous by Fundamental theorem of calculus. Is my reasoning correct? Any help?

$$|F(x)| \leq \int_{-\infty}^{\infty} |f(t)|\, dt$$ so $$F$$ is bounded. It is uniformly continuous because integrability of $$f$$ implies that for every $$\epsilon >0$$ there exits $$\delta >0$$ such that $$\int_E |f(t)| \, dt <\epsilon$$ whenever Lebesgue measure of $$E$$ is less than $$\delta$$. [ Take $$E=(x,y)$$ to get uniform continuity].
$$f(x)=\frac 1 {\sqrt x}$$ for $$0 and $$0$$ for other $$x$$ gives an example where $$f$$ is integrable but not bounded. Your example of a discontinuous integrable function $$f$$ is OK but you can make it much simpler. Take $$f=I_{(0,1)}$$ for example.
• $e^{-x^{2}}$ is bounded. Its values are all between $0$ and $1$. The other example is OK. – Kavi Rama Murthy Feb 28 '19 at 9:57