Show that
$ f(x) = \begin{cases} x^{\frac{-1}{2}}\sin(\frac{1}{x}) , & 0 <x\leq 1 \\
0 , & x=0 \end{cases}$
is integrable on $[0,1]$. Integrable means that $\displaystyle\int_{[0,1]} |f(x)|\,\mathrm dx < \infty$.
My attempt:
First, I checked to see if $f(x)$ had a bounded derivative on $[0,1]$. If it did, then it would be a Lipschitz function, hence absolutely continuous and hence integrable. Unfortunately, the derivative is not bounded.
Okay, so then I noticed that if $0 < a \leq 1$ then $f(x)$ is a well defined continuous function and so Riemann integrable on $[0,1]$, and hence Lebesgue integrable on $[0,1]$.
Next, I need to figure out how to prove that $f$ is continuous. I wouldn't mind some help with this part. Then I have that $\lim_{a \rightarrow 0^+} f(a)=f(0)$.
So, $$f(1)-f(0) = \lim_{a \rightarrow 0^+} f(1)-f(a) = \lim_{a \rightarrow 0^+} \int_{[a,1]} f(x)\,\mathrm dx = \lim_{a \rightarrow 0^+} \int_a^1 f(x)\,\mathrm dx.$$
Ugh idk honestly I'm lost. Would appreciate help. I'm pretty slow so lots of details appreciated!
Best Answer
You can just do it like $\displaystyle\int_{0}^{1}|f(x)|\,\mathrm dx\leq\int_{0}^{1}x^{-1/2}\,\mathrm dx=2<\infty$.