Is $\displaystyle f(x)=\sqrt{-2\ln{x}}$, $x \in (0,1)$ (inverse of the Gaussian curve) Lesbegue integrable ?
If not, is there any function that tends to $\infty$ at some point $x$ and is still integrable ? If yes, can you give me a sequence of simple functions that converge to $f$ ?
Your $f(x)$ is indeed Lebesgue integrable, and has integral $\sqrt{\pi/2}$.
However, if you find the series of simple functions in the above case too unweildy, try something simpler: $f: (0,1] \to \mathbb{R}$, $f(x) = n$ whenever $\frac{1}{(n+1)^2} < x \le \frac{1}{n^2}$.
Consider $x\mapsto\sqrt x$ over $(0,1)$. Apply the usual construction of Lebsgue, i.e. taking horizontal slices over $(0,1/n)$, and over $(0,1/n)$ take the simple function $\sqrt n\chi_{(0,1/n)}$.
