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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 $$f(x)=\begin{cases} 0 &\text{if}\;-\infty<x<0\\\\\text{linear}&\text{if}\;0\leq x\leq1/2\\\\\text{linear}&\text{if}\;1/2\leq x\leq1\\\\1&\text{if}\;x=2\\\\0&\text{otherwise}. \end{cases} $$

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

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$|F(x)| \leq \int_{-\infty}^{\infty} |f(t)|\, dt$ 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 $0<x<1$ 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.

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  • $\begingroup$ Thanks! Is my counterexamples for others correct? $\endgroup$ – Chinnapparaj R Feb 28 '19 at 9:53
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    $\begingroup$ $e^{-x^{2}}$ is bounded. Its values are all between $0$ and $1$. The other example is OK. $\endgroup$ – Kavi Rama Murthy Feb 28 '19 at 9:57

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