Show that the function is Lebesgue measurable

Let $f:[0,1] \to \mathbb R$ be Lebesgue integrable. Assume that $f$ is differentiable at $x=0$ and $f(0)=0$. Show that the function $g:[0,1] \to\mathbb R$ defined by $g(x)=x^{-3/2}f(x)$ for $x\in (0,1]$ and $g(0)=0$ is Lebesgue integrable.

I am studying for finals and am trying to wrap my head around this question. Would you use the Lebesgue-Vitali theorem that states, A bounded function $f:[a,b] \to\mathbb R$ is Riemann integrable if and only if it is continuous almost everywhere? or is this problem dealing with partitions, more specifically the mesh?? any help would be appreciated

• or here's another thinking process. I found a theorem that states, Let f be a measurable function. If there exists two integrable functions h and g s.t. h<=f<=g almost everywhere, then f is also an integrable function. If I can prove that g(x) is integrable over [0,1] along with this function, does that finish the proof? – cele Dec 9 '13 at 21:55
• I suggest dividing the proof into two; you should show that (a) $g$ is measurable, and (b) $\int_{[0,1]}|g|d\lambda<\infty$. – Jonathan Y. Dec 9 '13 at 21:58

We know $g$ is measurable because the functions $x \mapsto x^{-{3 \over 2} } 1_{(0,1]}(x)$ and $f$ are.
Since $f$ is differentiable at $x=0$ and $f(0) = 0$, we know that for some $K$ and $\delta>0$ we have $|f(x)| = |f(x)-f(0)| \le K |x-0| = K|x|$ for any $x \in [0,\delta)$.
Hence we have $|g(x)| \le |x^{-{3 \over 2} } | K |x|= K {1\over \sqrt{|x|}}$ for all $x \in (0,\delta)$.
Furthermore, since $x \mapsto x^{-{3 \over 2} }$ is decreasing on $(0,1]$, we have $0\le |g(x)| \le ({\delta \over 2})^{-{3 \over 2}} |f(x)|$ for all $x \ge {\delta \over 2}$.
Combining, we see that \begin{eqnarray} \int |g| &\le& \int_{[0,{\delta \over 2})}|g| + \int_{[{\delta \over 2},1}|g| \\ &\le& \int_0^{\delta \over 2}K {1\over \sqrt{x}} dx + ({\delta \over 2})^{-{3 \over 2}} \int_{[{\delta \over 2},1} |f| \\ &=& \sqrt{\delta \over 2}K+{\delta \over 2}^{-{3 \over 2}} \int |f| \end{eqnarray} and so $g$ is integrable.