Let $f$ be integrable over $\mathbb{R}$. Show that the function $f$ defined by $$F(x)=\int_{-\infty}^x f,\ \ \text{for all }\ x\in \mathbb{R}$$
is propely defined and continious. Is it necessarily Lipschitz?
First, since $f$ is integrable over $\mathbb{R}$, then $\int_\mathbb{R}|f|< \infty.$ Since $f$ is integrable over $\mathbb{R}$ and $C:=(-\infty,x]$ is measurable we have $\int_Cf=\int_\mathbb{R}f \chi_{_{C}}.$ Since $|f\chi_{_{C}}|\leq |f|$, we have $\int_{C}|f|=\int_\mathbb{R} |f\chi_{_{C}} |\leq \int_\mathbb{R}|f|< \infty$, so $\int_C f$ exists. Now, we see clearly that $F(x)=\int_{- \infty}^xf=\int_{-\infty}^yf=F(y)$ iff $x=y$, so $F$ is well defined function. Now we show that $F$ is continuous on $\mathbb{R}.$ Let $x \in \mathbb{R}$, and let $\epsilon >0$ be given. We see first that $\{(-n,x)\}_{n \geq 1}$ is a countable collection of an ascending sequence, so by the countinuity of integration we have that $$F(x)=\int_{- \infty}^xf=\int_{\bigcup_{n \geq 1}[-n,x]}f=\lim_{n \rightarrow \infty}\int_{-n}^xf.$$
The next part I am not confident about it:
Now for our $x$, there exists $\delta_1$ such that $\int_{-n}^{x}|f|<\epsilon/2$ whenever $x+n<\delta_1$. Also, for $y \in \mathbb{R}$, there exists $\delta_2$ such that $\int_{-n}^{y}|f|< \frac{\epsilon}{2}$ whenever $y+n<\delta_2$. Now, choose $\delta=\delta_1+\delta_2,$ if $|x-y|<\delta$, then we have $$|F(x)-F(y)|=|\lim_{n \rightarrow \infty}[\int_{-n}^xf-\int_{-n}^yf]|\leq \lim_{n \rightarrow \infty}\int_{-n}^x |f|+\lim_{n \rightarrow \infty}\int_{-n}^{y}|f|=\lim_{n \rightarrow \infty}[\int_{-n}^{x}|f|+\int_{-n}^y|f|]$$
$$<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
Hence, we conclude that for any $y \in \mathbb{R}$ such that $|x-y|<\delta$, then $|F(x)-F(y)|<\epsilon,$ so $F$ is continuous at $x$, hence on $\mathbb{R}.$
Also, I do not know how to show if it is necessary Lipschitz or not, so I would appreciate any help with that.
Best Answer
Take $f:\mathbb{R} \to \mathbb{R} $ defined as follows $$f(t)=\begin{cases} 0 \hspace{1.85cm}\mbox{ for } t\leqslant 0 \\ (1-\alpha ) t^{-\alpha } \mbox{ for } 0<t<1 \\0 \hspace{1.85cm}\mbox{ for } t\geqslant 1\end{cases}$$ for some $0<\alpha <1$. Then $f$ is integrable since $$\int_{\mathbb{R}} f(t) dt =\int_{(0,1 )} f(t) dt = t^{1-\alpha} |_0^1 =1.$$ For $0<u<v<1$ we get $F(v)-F(u) =v^{1-\alpha } -u^{1-\alpha }.$ Suposse that there exists $M>0 $ such that $$|F(v)-F(u)|\leqslant M|v-u|$$ then $$\frac{F\left(\frac{1}{n}\right) -F\left(\frac{1}{2n}\right)}{\frac{1}{n} -\frac{1}{2n}} \leqslant M$$ for all $n\in\mathbb{N} $ but this implies that $$\frac{n^{\alpha -1} -2^{\alpha -1} n^{\alpha -1}}{2^{-1} n^{-1} } \leqslant M$$ for all $n\in\mathbb{N}.$ Therefore $$(2-2^{\alpha} )n^{\alpha} \leqslant M$$ for all $n\in\mathbb{N}.$ But the last inequality is obviously not true.