Is the function $F(x)=\int _{-\infty}^x f(t) dt$ uniformly continuous?

Question

Let $f:\mathbb R\to \mathbb R$ be such that $\int _{-\infty}^{\infty} |f(x)|dx<\infty $.

Define $F:\mathbb R\to \mathbb R$ by $F(x)=\int _{-\infty}^x f(t) dt$

Does it follow that :

1. $f$ is continuous?
2. $F$ is uniformly continuous?
3. $|f|<M$ for some $M>0$?

I think $1$ is false since we can take $$f(x) = \begin{cases} 1, & \text{if $x\in [0,1]$ } \\ 0, & \text{elsewhere} \end{cases}$$

$2$ is true since $|F(x_1)-F(x_2)|=|\int _{x_1}^{x_2} f(t)dt| \leq \int _{x_1}^{x_2}| f(t)|dt$

**Since $\int _{-\infty}^{\infty} |f(x)|dx<\infty \implies |f(x)|\leq M.$

Is the statement marked with ** true as that is required to prove $2,3$?

Answer 1

Let $f(x)=\left\{\begin{matrix} x,x\in\mathbb{N}\\ 0,x\not\in\mathbb{N} \end{matrix}\right.$

So $\int_{-\infty }^{+\infty}f(x).dx=0$, but $f$ does not have supremum in $\mathbb{R}$, therefore the statement 3 is false.

Answer 2

Lets get everything together. I assume that all integrals are Riemann integrals. Recall basic facts:

A. When $f$ has countable number of discontinuities on $[a,b]$ then it is R-integrable on this interval.

B. The condition $\int_{- \infty}^\infty |f(x)| \mathrm{d} x < \infty$ means that both lmits $\lim_{b \rightarrow \infty} \int_0^b |f(x)| \mathrm{d} x $ and $\lim_{b \rightarrow -\infty} \int_b^0 |f(x)| \mathrm{d} x $ exist in finite sense

C. The function $ F(x) = \int_{-\infty}^x f(t) \mathrm{d} t$ makes sense, is continuous and differentiable at points of continuity of $f$

Hence:

Ad 1,3. No, need not be, you already have example that contradicts it.

Ad 2. Yes, it is as a continuous function with finite limits at infinities. To prove it you take interval $[-M,M]$ outside of which you have control over functions behavior due to finite limits. Inside this interval the function is uniformly continuous. Take this two together and you have overall uniform continuity.