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Prove that if $f$ is Riemann integrable on $[a,b]$ then so is $f^2$.

I have already proved that a function is Riemann integrable if and only if it is bounded and continuous a.e. If $f$ is bounded, then so is $f^2$. If $f$ is continuous a.e. then so is $f^2$ because it is a composition of a continuous function and a function that is continuous a.e.

But what if I was asked to prove the proposition directly (without refering to the above theorem)? Is that easy or technical?

Lemma: If function $f:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable then it is bounded on $[a,b]$.

Proof:

It is clear that $f$ is integrable if and only if for every $\epsilon>0$ there is $\delta>0$ such that $|S_1-S_2|<\epsilon$ whenever $S_1$ and $S_2$ are Riemann sums corresponding to partitions of $[a,b]$ of diameter less than $\delta$.

Choose $\epsilon>0$ and aparition of $[a,b]$ such that for arbitrary $x_i^{'}, x_i^{''} \in [x_{i-1},x_i]$, $i=1,\ldots, N$ we have

$$\Bigg|\sum_{i=1}^n (f(x_i^{'})-f(x_i^{''}))(x_{i}-x_{i-1}) \Bigg |< \epsilon$$

If we apply this inequality to the special case where, for some fixed index $j=1,\ldots, N$, we have $x_i^{'}= x_i^{''}$ if $i\ne j$ and $x_j^{''}= x_j$, we get

$$|(f(x_j^{'})-f(x_j))(x_{j}-x_{j-1})|<\epsilon$$ implying $$|f(x_j^{'})|<\frac{\epsilon}{x_{j}-x_{j-1}}+|f(x_j)|$$

This last inequality holds for all $x_j^{'}\in [x_{j-1},x_j]$, thus $f$ is bounded on $[x_{j-1},x_j]$. Therefore $f$ is bounded on all $[a,b]$.

luka5z
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    See this – Pedro Dec 22 '15 at 15:05
  • Is it really true that Riemann integrability requires boundedness? I don't think that's true. – MPW Dec 22 '15 at 15:08
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    @MPW: I believe that boundedness is required for a Riemann integral on a closed interval. Improper integrals are another story. – Rory Daulton Dec 22 '15 at 15:09
  • @RoryDaulton Only if $f$ is continuous, otherwise it needn't be bounded to be Riemann integrable. Consider $f$ on $[0,1]$ defined by $f(x)=1/\sqrt{x}$ for $x\not=0$ and $f(0)=0$. – Gregory Grant Dec 22 '15 at 15:12
  • Perhaps you should post your proof that Reimann integrable implies bounded and continuous. That proof must be wrong so if you post it we can look for the mistake together. – Gregory Grant Dec 22 '15 at 15:23
  • @GregoryGrant I didnt say it implies continuity, but only continuity almost everywhere (a.e.) – luka5z Dec 22 '15 at 15:27
  • @luka5z Ok I see that. Then please post the proof that it is bounded because that has to have a mistake in it. – Gregory Grant Dec 22 '15 at 15:29
  • @GregoryGrant I have edited my post. – luka5z Dec 22 '15 at 15:41
  • @luka5z If you do not allow improper integrals in your definition of Riemann integrable then that's okay. But you should specify that because some treatments will call an improper integral Riemann integrable. – Gregory Grant Dec 22 '15 at 15:43
  • @luka5z, sorry about that. I just posted a problem from my book. – luka5z Dec 22 '15 at 15:46
  • @GregoryGrant $f(x)=\begin{cases} 1/\sqrt{x} & x \in (0,1] \ 0 & x=0 \end{cases}$ is not Riemann integrable. It is improperly Riemann integrable, however. – Ian Dec 22 '15 at 15:48
  • @Ian I don't believe your definition is universal. See my comment below. – Gregory Grant Dec 22 '15 at 15:49
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    Here and here there are good discussions and references on the "fact" that Riemann integrability implies boundness – Pedro Dec 22 '15 at 15:57

2 Answers2

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Since $f$ is Riemann integrable on $[a,b]$ there is an $M$ with $|f(x)|\leq M$ for all $x\in[a,b]$. The function $g(x):=f^2(x)$ then satisfies $$|g(x)-g(y)|=|f(x)+f(y)|\>|f(x)-f(y)|\leq 2M \>|f(x)-f(y)|$$ for arbitrary $x$, $y\in[a,b]$. This implies that any test you can think of to establish the integrability of $f$ will also be passed by $g$.

Christian Blatter
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Consider the counter-example: $f$ on $[0,1]$ defined by $f(x)=1/\sqrt{x}$ for $x\not=0$ and $f(0)=0$. In this case $f$ is Riemann integrable but $f^2$ is not.

Gregory Grant
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  • This function is not Riemann integrable...or am I missing something? $f$ is not bounded. Every Riemann integrable function has to be bounded. – luka5z Dec 22 '15 at 15:20
  • @luka5z Yes it is. Just like $\int_1^{\infty}\frac{1}{x^2}dx$ is. – Gregory Grant Dec 22 '15 at 15:21
  • @luka5z A function does not have to be bounded to be Riemann integrable. Perhaps you want to add to your assumptions that $f$ is continuous. In that case it is true. – Gregory Grant Dec 22 '15 at 15:22
  • But I can prove that if function $f$ is integrable on $[a,b]$ then it is bounded. It's an easy proof. Are you interested in seeing it? – luka5z Dec 22 '15 at 15:25
  • @luka5z Yes please post that proof. – Gregory Grant Dec 22 '15 at 15:25
  • I will in 15 minutes and will edit my answer. – luka5z Dec 22 '15 at 15:26
  • Please see this post, I think this is the correct statement where you have to assume boundedness first: http://math.stackexchange.com/questions/972399/proof-that-if-f-is-integrable-then-also-f2-is-integrable – Gregory Grant Dec 22 '15 at 15:27
  • A function is Riemann integrable if the Riemann sums on arbitary fine partitions converge, this implies that the function must be bounded. Generaised integrals are limits of Riemann integrals, not Riemann integrals themselves. – mlu Dec 22 '15 at 15:32
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    Anyone to complete the following? If $f$ is Riemann integrable on $[a,b]$, then for every $x\in[a,b]$ there are step functions $s$ and $t$ such that $s\le f\le t$. And, for any given $\varepsilon>0$ there are corresponding $t$ and $s$ step functions such that

    $$\int_a^b t-\int_a^b s<\varepsilon$$

    We know that $s^2\le f^2\le t^2$ or $t^2\le f^2\le s^2$ is satisfied and know that $s^2$ and $t^2$ are also step functions. Assume that the first one is satisfied. Then...

     – frosh Dec 22 '15 at 15:39
  • I think our disagreement here is with definitions. If you do not allow improper integrals in your definition of Riemann integrable then you can conclude this, but you need to be clear about that because it's ambiguous in textbooks. – Gregory Grant Dec 22 '15 at 15:41
  • @Gregory so what is your defintion of Riemann integrability?? – mlu Dec 22 '15 at 15:43
  • @mlu I always took convergent improper integrals which are limits of Riemann integrable functions that have finite limit, to be Riemann integrable. I'm pretty sure my graduate analysis textbook (Royden) defined it that way, but I'm travelling so I can't check right now. But I'm almost certain in Rodyen there's an example of a function that's Riemann integrable but not Lebesgue integrable, that assumes this. – Gregory Grant Dec 22 '15 at 15:47
  • I've always heard Riemann integrability to imply proper Riemann integrability. Without this you get awkward things, for instance "Riemann integrability implies Lebesgue integrability" is only true if you mean "proper Riemann integrability". In Royden & Fitzpatrick, the very first sentence that is stated about the Riemann integral explicitly states that the functions in question are bounded and defined on a closed interval. The definition in chapter 4 also assumes boundedness. – Ian Dec 22 '15 at 15:52
  • Riemann integral definition is on bounded functions. – frosh Dec 22 '15 at 15:53
  • @Ian Fitzpatrick massively changed Royden, go back to the last edition of Royden before Fitzy got his hands on it and I believe there is an example of a function that is Riemann but not Lebesgue integrable that relies on the definition of R.I. including improper integrals. – Gregory Grant Dec 22 '15 at 15:54
  • Ok we've clearly gotten to the bottom of this. All disagreements in this thread relate to definitions and not proofs. I'm satisfied to move on. – Gregory Grant Dec 22 '15 at 15:55
  • OK, well, consider the theorem that Riemann and Darboux integration are equivalent. If you have a function which is unbounded above, any of its upper sums are $+\infty$. How should such a thing possibly be construed as being Darboux integrable? And yes, everything here is about definitions...except I've never heard of a definition that says "Riemann integrable" and means "possibly improperly Riemann integrable". I've always seen the two situations explicitly distinguished, because it drastically muddles the situation to do the opposite. – Ian Dec 22 '15 at 15:55