Using the monotone convergence theorem to show a function is integrable

Question

Apply the monotone convergence theorem and the fundamental theorem of calculus to show that

$f(x) = \left\{ \begin{array}{ll} x^{-a} & \mbox{if } 0 < x \leq 1 \\ \infty & \mbox{if } x = 0 \end{array} \right. $

is integrable if and only if $a < 1$.

Hint: For $0 \leq a \leq 1$, set

$f_n(x) = \left\{ \begin{array}{ll} x^{-a} & \mbox{if } {\frac{1}{n}} \leq x \leq 1 \\ n^a & \mbox{if } 0 \leq x < \frac{1}{n} \end{array} \right. $

So far, I can show that $f_n \nearrow f$ (the criteria for using MCT), but dont know where exactly to use MCT. Also, how/where the fundamental theorem of calculus comes into it. Any help/thoughts would be appreciated. Thanks.

Answer 1

So the MCT tells you that $\int f=\lim_n\int f_n$ you can then use the FTC to compute the RHS which will be $\lim_{n\rightarrow\infty}n^a\cdot \frac 1 n+ \frac{x^{-a+1}}{1-a}\vert^1_{\frac 1 n}$ which is finite if and only if $a<1$.