$f$ is a bounded function, $f:[a,b]\rightarrow R$ Prove that if $f$ is continuous on (a,b], then $f$ is integrable on [a,b]

Question

My question is: Is there any way to prove this other than using a $\delta-\epsilon$ with Darboux/Riemann sums? And if so, could I get a hint on what to use to fulfill this proof?

I don't think this is true - take $f(x)=0$ for $x=0$, $f(x)=1/x$ for $x\in (0,1]$. — Wojowu, Jun 12 '16 at 17:29
@Wojowu My initial edit seemed to have taken out a good portion of the question. If $f$ is bounded it seems that the above can be proved. — RonaldB, Jun 12 '16 at 17:44
@ParamanandSingh haven't touched Riemann Integrals as of yet. — RonaldB, Jun 12 '16 at 20:19
I think you need to study Riemann integrals then. Without a proper theory of Riemann integral how do you define the symbol $\int_{a}^{b}f(x),dx$? If you are defining it as $F(b) - F(a)$ where $F$ is an anti-derivative of $f$ then you are restricting yourself to integration of functions continuous on $[a, b]$. — Paramanand Singh, Jun 12 '16 at 20:23
@ParamanandSingh it was defined using Darboux sums. I'm not too sure of the difference/similarities between them though. — RonaldB, Jun 12 '16 at 20:24
Good then you are already there. A Riemann sum is always sandwiched between lower and upper darboux sums and only very little effort is needed to understand Riemann sums. I suggest you read the answer given for the linked question. — Paramanand Singh, Jun 12 '16 at 20:27
@ParamanandSingh is one more useful than the other? Riemann sums/integrals seem to be much more popular than Darboux on here. — RonaldB, Jun 12 '16 at 20:30
Riemann and Darboux integrals are equivalent concepts and anyone studying integrals should know both of them. Some of the proofs are easier via one approach and some other proofs require another approach. — Paramanand Singh, Jun 12 '16 at 20:34

score 3 · Answer 1 · edited Apr 13 '17 at 12:21

A proper proof does not seem to be possible without the use of $\epsilon, \delta$ arguments. I will try to provide a proof using Darboux sums.

First we note the following condition of integrability:

A function $f$ bounded on $[a, b]$ is Riemann integrable on $[a, b]$ if and only if for any $\epsilon > 0$ there is a partition $P$ of $[a, b]$ such that $$U(P, f) - L(P, f) < \epsilon$$ where $L(P, f), U(P, f)$ are lower and upper darboux sums for $f$ over partition $P$.

We are given that $f$ is bounded on $[a, b]$ and thus let $M, m$ be the supremum and infimum of $f$ on $[a, b]$. Now given any $\epsilon > 0$ it is possible to choose a positive number $h$ such that $(M - m)h < \epsilon / 2$. Also let $h$ be small enough to satisfy $h < b - a$ and consider $c = a + h$. We are given that $f$ is continuous on $(a, b]$ and hence $f$ is continuous on $[c, b]$ and therefore Riemann integrable on $[c, b]$. Thus there is a partition $P'$ of $[c, b]$ such that $$U(P', f) - L(P', f) < \frac{\epsilon}{2}$$ Consider the partition $P$ of $[a, b]$ obtained by adding point $a$ in $P'$ i.e. $P = P' \cup \{a\}$. We can see that if $M', m'$ are supremum and infimum of $f$ on $[a, c]$ then $$U(P, f) - L(P, f) = (M' - m')(c - a) + U(P', f) - L(P', f)$$ Noting that $c = a + h$ and the fact that $M' - m' \leq M - m$ we get $$U(P, f) - L(P, f) < (M - m)h + \frac{\epsilon}{2} < \epsilon $$ It now follows that $f$ is Riemann integrable over $[a, b]$.

What we see here is that discontinuity at possibly an end point of the interval does not impact the integrability of $f$. By splitting interval $[a, b]$ into a finite number of sub-intervals it is possible to extend the above result to get the following:

If $f$ is bounded on $[a, b]$ and $f$ is continuous except for a finite number of discontinuities in $[a, b]$ then $f$ is Riemann integrable on $[a, b]$.

However this is not the final story. It is possible that $f$ can have infinite number of discontinuities and yet be integrable (for example monotone functions are Riemann integrable but can possess at max a countable number of discontinuities, also see this answer). There are functions which have an uncountable number of discontinuities and are yet Riemann integrable. In fact counting the number of discontinuities of $f$ to infer Riemann integrability is not the right approach, but rather one needs to figure out how much space these points of discontinuities take up from the interval $[a, b]$ and this view was championed by Henri Lebesgue to develop his theory of measure and integration.

score 0 · Answer 2 · answered Jun 12 '16 at 17:56

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This is a simple consequence of the Lebesgue's theorem:

Bounded function $f\colon [a,b]\to \mathbb R$ is R-integrable if and only if it is continuous almost everywhere (in the sense of Lebesgue measure)

Since you have a bounded function on $[a,b]$, which is continuous everywhere except at $\{a\}$, R-interability is obvious by the above theorem.

answered Jun 12 '16 at 17:56

Ennar

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It's definitely not an adapted answer ! – Surb Jun 12 '16 at 19:11
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@Surb, "My question is: Is there any way to prove this other than using a δ−ϵ with Darboux/Riemann sums?" Of course, any statement about Riemann integral must use it's definition directly or indirectly, but my answer fits the requirements. – Ennar Jun 12 '16 at 19:32
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The problem with this is that proving Lebesgue's theorem is much more difficult than solving the given problem. – MathematicsStudent1122 Jun 12 '16 at 19:34
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@MathematicsStudent1122, definitely. But nevertheless, I'd appreciate if someone gave me a general result that directly applies to given problem, especially if the result is not from obscure paper, but something that will be encountered in near future. Opinions differ, of course. – Ennar Jun 12 '16 at 19:42
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@Ennar: "I'd appreciate if someone gave me a general result that directly applies to given problem". I agree if you can understand the general result. If not, it's definitely inappropriate. And for the OP, it's not the case ! – Surb Jun 12 '16 at 19:55
@Surb, as I said, opinions differ. Personally, I find it more inappropriate to downvote something that OP will encounter in the next year, year and a half as if it were inaccurate, and leave an impression that this does not answer the question correctly. – Ennar Jun 12 '16 at 20:03
@Ennar Agreed. Downvotes are absurd, because the answer is not wrong. – MathematicsStudent1122 Jun 12 '16 at 22:23

$f$ is a bounded function, $f:[a,b]\rightarrow R$ Prove that if $f$ is continuous on (a,b], then $f$ is integrable on [a,b]

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