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Let $\sigma: [ 0,\infty) \rightarrow [0,\infty]$ be a Borel-measurable, nonnegative, extended-real valued function, when is the Lebesgue integral $\int_{0}^{t}\sigma(u) du$ a continuous function of t? Thanks!

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  • $\begingroup$ Note that any antiderivative of a locally integrable function is not merely continuous, but also absolutely continuous. $\endgroup$ – Sangchul Lee Jul 28 '12 at 17:34
  • $\begingroup$ Do we know something else, like $\sigma\in L^1_{loc}[0,\infty)$ or something? $\endgroup$ – Matt Jul 28 '12 at 17:40
  • $\begingroup$ If $\sigma$ is locally integrable, then the function is in fact differentiable a.e. (ie, the Lebesgue points). $\endgroup$ – copper.hat Jul 28 '12 at 17:42
  • $\begingroup$ Notice that $\sigma$ is nondecreasing, which implies that the set $\{t: \sigma(t)<\infty \}$ is an interval. In this interval (possibly empty) you have local integrability, hence local absolute continuity of the antiderivative. Outside of it, the antiderivative is $\equiv \infty$. $\endgroup$ – user31373 Jul 28 '12 at 19:08
  • $\begingroup$ @Leonid: No assumption on the variations of $\sigma$. $\endgroup$ – Did Jul 28 '12 at 19:43
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Since my answer-comment had a couple of typos and did not cover all bases, I'm posting an actual answer in the answer box (what a novel idea!).

Let $F(t)=\int_0^{t}\sigma(u)\,du$, which is a well-defined function taking values in $[0,\infty]$. Since $\sigma\ge 0$, $F$ is nondecreasing, which implies that the set $I=\{t:F(t)<\infty\}$ is an interval. On this (possibly empty) interval $\sigma$ is locally integrable, and therefore $F$ is locally absolute continuous. Outside of the interval $F\equiv +\infty$.

The question remains: what happens at right endpoint of $I$ when this endpoint is finite? Here a discontinuity may occur, e.g., if $\sigma\equiv \infty$ or $\sigma=\infty \chi_{[1,\infty)}$ or, to give a finite example, $\sigma(u)=(u-1)^{-1}\chi_{(1,\infty)}$. Denoting the right endpoint of $I$ by $b$, we can formulate the criterion for continuity at $b$ (hence on the entire line) as $\int_0^b \sigma(u)\,du=\infty$.

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  • $\begingroup$ Thank you for this answer, it is very clear. $\endgroup$ – Jean Carr Aug 18 '12 at 0:15

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