Real Analysis: if the integral of the cube of a function exists, does it follow that the integral of the function also exists? Let $I=[a,b]$. Given that $\int_a^bf^3(x)\;dx$ exists, does it follow that  $\int_a^bf(x)\;dx$ exists?
Let's let $a$ and $b$ be real numbers.
 A: Hint (for a solution not using measure theory).
If you know a statement that says: if $g$ is continuous and $f$ is Riemann-integrable then $g \circ f$ is Riemann-integrable, you're done!
If not, you can try to prove it with following steps:


*

*If $f$ is Riemann-integrable and $g$ Lipschitz then $g \circ f$ is Riemann-integrable.

*A continuous function defined on a compact segment $[a,b]$ is the uniform limit of a sequence of piecewise linear functions.

*A piecewise linear function is Lipschitz.

*Conclude based on 1, 2 and 3.


You can simplify a bit this general statement for your $g: x \to x^{1/3}$, but I think that it is more valuable for you to prove the general statement as it let you understand deeper what is happening.
A: The claim is true if $[a,b]$ is a finite interval. We may assume without loss of generality that $f \geq 0$ on this interval, as the integral of measurable function exists only if its "positive" and "negative" parts do.
Let $S = \{ x \in [a,b] : f(x) \leq 1 \}$. Notice that for any $s \in S^C$, $ \, \,f(s) \leq f(s)^3$, since $f(s) > 1$. Then
$$
\int_{a}^{b} f(x) \, dx = \int_{S} f(x) \, dx + \int_{S^C} f(x) \, dx \leq  \int_{S} \, dx + \int_{S^C} f(x)^3 \, dx
$$
Note that here we have used the fact that on $S$, $f \leq 1$. But since $f^3$ is integrable on $[a,b]$ and $S \cup S^{C} = [a,b]$,
$$
\int_{a}^{b} f(x) \, dx \leq (b-a) + \int_{a}^{b} f(x)^3 \,dx < \infty.
$$
Thus $f$ is integrable on this domain.
(Sidebar: Here I've been using the Lebesgue integral, but provided $f$ is nice enough, (Super secret sidebar: I think just continuous) $S$ and $S^C$ should just be the union of intervals, so this would also work for the Riemann integral.)
A: Solution using "a gun to kill a fly" (straight translation of a French sentence).
According to Lebesgue criteria a function is Riemann integrable if and only if the measure of the set of its discontinuity points is zero. As $x \to x^3$ is continuous as its inverse function, the measure of the set discontinuity points of $f^3$ is the same as the one of $f$. Hence $f$ is Riemann integrable.
A: Here is an elementary  proof, using just the definition of Riemann integrability:
Write $f^3(x)=:g(x)$. We may assume $[a,b]=[0,1]$ and $\bigl|g(x)\bigr|\leq1$, since $g$ is Riemann integrable. Given any subinterval $J\subset[0,1]$ write $$\|\Delta g\|_J:=\sup_{x,y\in J}\bigl|g(y)-g(x)\bigr|\ ,$$ and denote the length of $J$ by $|J|$.
Let an $\epsilon>0$ be given. By assumption there is a partition of $[0,1]$ into finitely many subintervals $J_k$ such that
$$\sum_k\|\Delta g\|_{J_k}\>|J_k|\leq\epsilon^4\ .\tag{1}$$
Call a $k$  bad if $\|\Delta g\|_{J_k}\>\geq\epsilon^3$ and good otherwise. From $(1)$ it then follows that
$$\sum_{k\>{\rm bad}}|J_k|<\epsilon\ .$$
Looking at the graph of the function $t\mapsto{\root 3\of t}$ (defined on all of ${\mathbb R}$ and having a sign) we see that
$$\bigl|{\root 3\of u}-{\root 3\of v}\bigr|\leq2^{2/3}|u-v|^{1/3}\qquad(u,v\in{\mathbb R})\ .$$
From this we conclude that
$$\|\Delta f\|_{J_k}\leq2^{2/3}\|\Delta g\|_{J_k}^{1/3}<2\epsilon\qquad(k\ {\rm good})\ ,$$
so that we obtain
$$\sum_k \|\Delta f\|_{J_k}\>|J_k|=\sum_{k\>{\rm good}} \|\Delta f\|_{J_k}\>|J_k|+\sum_{k\>{\rm bad}} \|\Delta f\|_{J_k}\>|J_k|\leq2\epsilon\cdot 1+2\cdot\epsilon=4\epsilon\ .$$
As $\epsilon>0$ was arbitrary this proves that $f$ is Riemann integrable over $[0,1]$
A: Basic result: If $f$ is Riemann integrable on $[a,b]$ and $g$ is continuous on $\mathbb R,$ then $g\circ f$ is Riemann integrable on $[a,b].$ Apply this to your problem with $g(x) = x^{1/3}.$
