Finding Lebesgue Integral of $\frac{1}{\sqrt{x}}$ over $(0,1]$ How do I rigorously discover what
$$
\int_{(0,1]} \frac{1}{x^{1/2}} = \underset{0 \le \phi \le \frac{1}{\sqrt{x}}}{\sup} \int_{(0,1]} \phi
$$
(for $\phi$ a simple function) is?  Note that I have already shown in another exercise that such an integral exists and that it is finite.
 A: Here is a brute force approach using the dominated convergence theorem (DCT). The idea is to create a sequence of simple functions $s_n$ such that $s_n(x) \leq \frac{1}{\sqrt{x}}$, and $s_n(x) \to \frac{1}{\sqrt{x}}$. The DCT shows that $\int s_n \to \int \frac{1}{\sqrt{x}}$.
Let $s_n = \sum_{k=n}^{n^2-1} \frac{k}{n} 1_{[(\frac{n}{k+1})^2, (\frac{n}{k})^2)}$. Then $s_n(x) \leq \frac{1}{\sqrt{x}}$,  $s_n(x) \to \frac{1}{\sqrt{x}}$, and $s_n$ is supported in $[\frac{1}{n^2},1)$.
Then
\begin{eqnarray}
\int s_n & = &  \sum_{k=n}^{n^2-1} \frac kn\left(\left(\frac{n}{k}\right)^2 - \left(\frac{n}{k+1}\right)^2\right) \\
& = & n \sum_{k=n}^{n^2-1}\left(\frac{1}{k} - \frac{k}{(k+1)^2}\right) \\
& = & n \sum_{k=n}^{n^2-1}\left(\frac{1}{k} - \frac{1}{k+1} +\frac{1}{(k+1)^2}\right) \\
& = & n\left(\frac{1}{n}-\frac{1}{n^2}\right) + n \sum_{k=n}^{n^2-1} \frac{1}{(k+1)^2}.
\end{eqnarray}
Noting that $\int_{k+1}^{k+2} \frac{dx}{x^2} \leq \frac{1}{(k+1)^2} \leq \int_k^{k+1} \frac{dx}{x^2}$, we obtain the bounds $\frac{1}{n+1}-\frac{1}{n^2+1} \leq \sum_{k=n}^{n^2-1} \frac{1}{(k+1)^2} \leq \frac{1}{n}-\frac{1}{n^2}$. It follows that $\lim_{n \to \infty} \int s_n = 2$, which coincides with the value obtained from the improper Riemann integral.
In terms of the original question, we note that if $s$ is simple, and $s(x) \leq \frac{1}{\sqrt{x}}$, then $\int s \leq \int \frac{1}{\sqrt{x}}$, and so the above shows that $\int \frac{1}{\sqrt{x}} = \sup_{s \, \mathbb{simple}, s(x) \leq \frac{1}{\sqrt{x}}} \int s = 2$.
