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Question

Suppose that f is bounded on $[0,b]$, and suppose $\int_{c}^b f $ exists for all $0<c<b$. Show that $\int_{0}^b f$ exists.

I first think that consider the interval $[b/n,b]$ for $n\ge2$.

Since integrable for $[b/n,b]$, if I denote the set $A_n$ as the set of discontinuities for $[b/n,b]$.

By Lebesgue's theorem, measure of $A_n$ is zero.

Since countable union of measure zero set has measure zero, I can integrable over [0,b].

But I think this approach is wrong since $\cup_{n=2}^{\infty}[b/n,b]=(0,b]$.

Is this really wrong? If not, I can tell that f is integrable over (0,b]?

help me to solve this problem.

Note that here the integration is Riemann integration, not Lebesgue integration.

zhw.
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Pearl
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    Note that you have proven that the set of discontinuities on $(0,b]$ has measure $0$. What could the measure of the set of discontinuities on $[0,b]$ possibly be? – Arthur Jun 01 '16 at 14:21
  • @Arthur Thanks for the comment! – Pearl Jun 01 '16 at 14:28

1 Answers1

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Alternate proof: Suppose $|f|\le M$ on $[0,b].$ Let $b>\epsilon>0,$ thinking of $\epsilon$ as small. Since $f$ is Riemann integrable on $[\epsilon,b],$ there is a partition $P$ of $[\epsilon,b]$ such that $U(P,f)-L(P,f)<\epsilon.$ It follows that for the partition $P\cup \{0\}$ of $[0,b],$ we have

$$\tag 1 U(P\cup \{0\},f) - L (P\cup \{0\},f) < \epsilon + 2M\epsilon.$$

The right side of $(1)$ can be made arbitrarily small, proving $f$ is Riemann integrable on $[0,b].$

zhw.
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