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Is any continuous bounded function on $(a,b)$ Riemann integrable? Traditionally, we don't discuss the Riemann integrability when the domain is not a closed and bounded interval. When dealing with other domain, such as $[a,\infty)$, we sometimes refer it to the improper integral. However, if we use the definition from the multivariable Riemann integration case, we can talk about such integral on $(a,b)$. So is any continuous bounded function on $(a,b)$ Riemann integrable under this meaning? If so, how can we prove it? Is their a way to avoid digging into $\epsilon-\delta$?

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    $\begingroup$ We can talk about Riemann integrability on $(a, b)$: just extend the function to $[a, b]$ by setting e.g. $f(a) = f(b) = 0$ (if $a, b$ are not infinty). Then every bounded continuous $f$ is Riemann integrable: see here. $\endgroup$ – user99914 Jan 2 '18 at 17:30
  • $\begingroup$ But is there a gap between $\overline{f}$ (means extension) is integrable and $f:(a,b)\to\Bbb R$ is integrable? Since the definition of a function on open interval (again notice that it is not an improper integral) via multivariable integration in $\Bbb R^n$ when $n=1$ is somehow weird. We've already know that $\overline{f}$ is integrable , should we prove $\overline{f}$ is integrable $\Longrightarrow f:(a,b)\to\Bbb R$ is integrable? $\endgroup$ – Eric Jan 2 '18 at 17:39
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Apostol in his Mathematical Analysis describes the multiple integrals on compact intervals (closed and bounded intervals which are product of many one dimensional closed intervals) and later extends the definition to Jordan measurable sets via the use of characteristic functions.

Thus the same argument applies to the one dimensional case also. The set $(a, b) $ is Jordan measurable and if $f$ is defined on $(a, b) $ and we create another function $g:[a, b] \to\mathbb{R} $ via $g(x) =f(x), x\in(a, b) $ and $g(a) =g(b) =0$ then we say that $f$ is Riemann integrable on $(a, b) $ if and only if $g$ is Riemann integrable on $[a, b] $ and then their integrals are equal.

It should now be obvious that if $f$ is bounded and continuous on $(a, b) $ then $f$ is Riemann integrable on $(a, b) $.


I don't know what is gained from the definition of multiple integrals given in Wikipedia which involves half open intervals. The approach by Apostol seems less clumsy and is general enough.

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A function is Riemann integrable on $[a,b]$ iff it is bounded and continuous save on a set of Lebesgue measure zero. Here the discontinuities of $f$ are within the finite set $\{a,b\}$ and so $f$ is Riemann integrable.

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  • $\begingroup$ Well.. but what if the original $f$ itself is defined on $(a,b)$? $\endgroup$ – Eric Jan 2 '18 at 17:20
  • $\begingroup$ It doesn't matter how one extends it to $[a,b]$. @Eric $\endgroup$ – Lord Shark the Unknown Jan 2 '18 at 17:21
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    $\begingroup$ First you should explain how you define integrability on an open interval… so when is $$\phi\vert_{(a,b)}$$ integrable $\endgroup$ – Gono Jan 2 '18 at 17:43
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    $\begingroup$ @Shashi: save means "except" here. $\endgroup$ – Paramanand Singh Jan 3 '18 at 2:53
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    $\begingroup$ @Eric the Riemann integral from $a$ to $b$ is the same as the integral of the RHS limit of $a$ to the LHS limit of $b$. So we can just extend $(a,b)$ to be $[a,b]$ as long as $a,b$ exist and get the same answer in the integral(it is actually the same integral) $\endgroup$ – Holo Jan 3 '18 at 5:03
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Riemann-Lebesgue Theorem states that f:[a,b]$\to$R is Reimann integrable if and only if f is bounded and the set of discontinuities is a zero set which could easily be applied to this case with probably discontinuities in x=a and x=b that are of course bounded since whole the function is so. In this case, the set of (probable) discontinuities include at most two points and then is a zero set.

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