Is the function $f(x)=\frac{1}{\sqrt{x}}$ Lebesgue integrable on $[0,1]$ ? Justify.
From intuition I believe it is Lebesgue integrable, since it is Riemann integrable. $$\int_0^1\frac{1}{\sqrt{x}}=\int_0^1 x^{-\frac{1}{2}}=2x^{\frac{1}{2}}\vert_0^1=2$$Also by then fact that the points of discontinuities is countable i.e$( x=o)$. Is there any formal way to show this results preferably by constructing simple function approximation of $f(x)$? Any help? thank you