I want to calculate the Lebesgue integral $\int_{[0,1]}x^{-3/2}d\lambda(x)$.
But like Riemann integration we do not have fundamental theorem of calculus as we have for Riemann integral so that we could say the Lebesgue integral equals $[-2x^{-1/2}]_0^1$. So I have only one candidate to check which is the improper Riemann integral $\int_0^1 x^{-3/2}dx=\lim\limits_{\epsilon\to 0+}\int_{\epsilon}^1x^{-3/2}dx$. Here this limit equals to $\lim\limits_{\epsilon\to 0+}(-2+\frac{2}{\sqrt \epsilon})=+\infty$.
So what can I conclude from here? Does the function $x^{-3/2}\in L^1([0,1])$ or not? I am confused. Can someone help me with this?
Based on the discussion in the comments,I am posting an answer to this question.Let, $f(x)=\begin{cases} x^{-3/2},\text{ if } 0<x\leq 1\\ 0,\text{ if } x=0\end{cases}$ which is measurable .Now consider the sequence of measurable functions $f_n(x)=\begin{cases} x^{-3/2},\text{ if } \frac{1}{n}\leq x\leq 1\\ 0,\text{ otherwise } \end{cases}$.Clearly $0\leq f_n\leq f_{n+1}$ and $f_n\to f$ pointwise.So,we can apply monotone convergence theorem and get, $\lim\limits_{n\to \infty}\int_{[0,1]} f_n(x)d\lambda(x)=\int_{[0,1]}f(x)d\lambda(x)\implies \lim\limits_{n\to \infty}\int_{[\frac{1}{n},1]}f(x)d\lambda(x)=\int_{[0,1]}f(x)d\lambda(x)$$\implies +\infty=\lim\limits_{n\to \infty}\int_{1/n}^1x^{-3/2}dx=\int_{[0,1]}f(x)d\lambda(x)$. So,$f(x)\notin L^1([0,1])$.