Prove that if $f$ is Riemann integrable on $[a,b]$ then so is $f^2$ 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 $\epsilon&gt;0$ there is $\delta&gt;0$ such that $|S_1-S_2|&lt;\epsilon$ whenever $S_1$ and $S_2$ are Riemann sums corresponding to partitions of $[a,b]$ of diameter less than $\delta$.

Choose $\epsilon&gt;0$ 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

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

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

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

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 $[a,b]$.
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
$$|g(x)-g(y)|=|f(x)+f(y)|\&gt;|f(x)-f(y)|\leq 2M \&gt;|f(x)-f(y)|$$
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.