Prove that $f$ is Riemann integrable iff $f$ squared is Riemann integrable for strictly positive

Question

Prove that the following are equivalent for non-negative function $f$ bounded on $[a,b]$
1. $f$ is riemann integrable
2. $f$ squared is riemann integrable

Answer 1

For $f$ nonnegative, the functions $f$ and $f^2$ have the same set of discontinuities. Moreover, a bounded function on a compact interval is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero.

Hence, $f$ is Riemann integrable if and only if $f^2$ is Riemann integrable [for $f$ nonnegative, as in the question].

Answer 2

Another approach is to use the fact that if $g$ is continuous on a closed interval containing the range of a Riemann integrable function $f$ on $[a,b],$ then $g\circ f$ is Riemann integrable on $[a,b].$ For the problem at hand, apply the above with $g(x) = x^2$ or $g(x) = \sqrt x$ to get the desired equivalence.

Answer 3

And if you want to avoid using Lebesgue criterion...

There is a theorem stating that for $f$ Riemann-integrable and $g$ continuous, $g \circ f$ is Riemann integrable.

From that:

1. If $f$ is Riemann integrable, $f^2 = h \circ f$ is integrable as $h : x \mapsto x^2$ is continuous.
2. If $f^2$ is Riemann integrable then $f = k \circ f$ is Riemann integrable as $k : x \mapsto \sqrt{x}$ is continuous and $f$ supposed to be non-negative.

Answer 4

To provide more details on the answer by Jonas, as requested by OP, I will show why $f$ and $f^2$ have the same sets of discontinuities.

Claim 1: Let $f\colon A\to B$ (not necessarily continuous) and $g\colon B\to C$ continuous function. Then:

$f$ is continuous at $x$ $\implies$ $g\circ f$ is continuous at $x$

This is straightforward from any definition of continuity, since $g$ is continuous at $f(x)$ by our assumption.

The converse doesn't hold in general, for example, one might take bounded $ f\colon[a,b]\to\mathbb R$ $$f(x)=\left\{ {\begin{array}{r l} 1, &x\in\mathbb Q\cap[a,b] \\ -1, &\text{otherwise} \end{array}}\right.$$ and continuous $g(x) = x^2$, in which case we have that $(g\circ f)(x) = 1$ is continuous, but $f$ isn't.

But, if we furthermore assume that $g$ is bijective, then the converse holds as well:

Claim 2: Let $f\colon A\to B$ (not necessarily continuous) and $g\colon B\to C$ continuous bijection. Then:

$f$ is continuous at $x$ $\iff$ $g\circ f$ is continuous at $x$

Proof. One direction comes from Claim 1. To get the other direction, apply Claim 1 on $g\circ f$ and $g^{-1}$, i.e. since $g\circ f$ is continuous at $x$, $g^{-1}\circ(g\circ f) = f$ is continuous at $x$ as well. $\square$

How does this prove that $f$ and $f^2$ have the same sets of discontinuities for nonnegative $f$? Well, $g\colon\mathbb R_{\geq 0}\to\mathbb R_{\geq 0}$, given by $g(x) = x^2$ is bijective and continuous, so $f^2 = g\circ f$ is continuous at $x$ if and only if $f$ is continuous at $x$, by our Claim 2.

And this is where we need to stress that $f$ being nonnegative is paramount since $x\mapsto x^2$ is bijective only on nonnegative reals, so $f$ must be nonnegative as well, otherwise we can't apply our Claim 2.