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 $[a,b]$.


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 $[a,b]$.

  • 2
    $\begingroup$ See this $\endgroup$ – Pedro Dec 22 '15 at 15:05
  • $\begingroup$ Is it really true that Riemann integrability requires boundedness? I don't think that's true. $\endgroup$ – MPW Dec 22 '15 at 15:08
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    $\begingroup$ @MPW: I believe that boundedness is required for a Riemann integral on a closed interval. Improper integrals are another story. $\endgroup$ – Rory Daulton Dec 22 '15 at 15:09
  • $\begingroup$ @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 $f(0)=0$. $\endgroup$ – Gregory Grant Dec 22 '15 at 15:12
  • $\begingroup$ 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. $\endgroup$ – Gregory Grant Dec 22 '15 at 15:23

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 $g$.


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.

  • $\begingroup$ This function is not Riemann integrable...or am I missing something? $f$ is not bounded. Every Riemann integrable function has to be bounded. $\endgroup$ – luka5z Dec 22 '15 at 15:20
  • $\begingroup$ @luka5z Yes it is. Just like $\int_1^{\infty}\frac{1}{x^2}dx$ is. $\endgroup$ – Gregory Grant Dec 22 '15 at 15:21
  • $\begingroup$ @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. $\endgroup$ – Gregory Grant Dec 22 '15 at 15:22
  • $\begingroup$ 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? $\endgroup$ – luka5z Dec 22 '15 at 15:25
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    $\begingroup$ 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... $\endgroup$ – frosh Dec 22 '15 at 15:39

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