Let $f$ be a function defined on [0, 1] that is integrable over [0, 1], differentiable at x = 0, with f(0) = 0. Define the piecewise function g(x) to be $x^{-3/2} f(x)$ if $0 < x \leq 1$ and 0 if $x = 0$. Prove $g$ is integrable over [0, 1].
The fact that the exponent on $x$ is less than -1 is tripping me up here. I know that in general, $x^{\alpha}$ is integrable over (0, 1] only when $\alpha > -1$, so if I had an exponent in that range, I would have a product of integrable functions. But as it is the integral diverges. I can't apply Holder's inequality either, since clearly $x^{-3/2}$ is not essentially bounded. Am I missing something, or might the problem have a typo in the exponent?
Putting this here since it's too long for a comment. This is only a rough outline of an approach, you have to fill in the details yourself.
I asssume you talk about Riemann integrability here. That means $f$ is bounded and by the Lebesgue criteria it is continous almost everywhere. Now use the fact's that $f(0)=0$ and $f$ is differentiable at $0$ to show that the function
$$ h(x)=\dfrac{f(x)}{x} $$ is integrable. Hint: It is bounded on $[0,1]$ since it is continous at $0$. And clearly it is continous almost everywhere since $f$ is. Now use Lebesgue's criteria again.
But now $g(x)=x^{-1/2}h(x)$, can you carry on from here?