Showing that $1/\sqrt{x}$ on $(0,1)$ is in $L^1$

Question

How can I go about showing that $\frac1{\sqrt{x}}$ on $(0,1)$, $0$ otherwise is in $\mathcal{L}^1(\mathbb{R})$? I've tried bounded convergence but I'm not sure what function to bound above by. For using monotone convergence lemma I've run into similar problems, especially around bounding the integrals of each function in my series. What's the elegant way to go about this?

Answer 1

One may observe that $$ \left(2\sqrt{x} \right)'=\frac1{\sqrt{x}}, \qquad x>0, $$ giving $$ \int_{\mathbb{R}}f(x)\:dx=\int_0^1\frac1{\sqrt{x}}\:dx=\left[2\frac{}{}\sqrt{x} \right]_0^1=2<\infty $$ thus the given function satisfies $$ f \in \mathcal{L}^1(\mathbb{R}). $$

Answer 2

Define $f_n=\chi _{(1/n,1)}\cdot f$ and use the Monotone Convergence Theorem:

$\int f(x)dx=\lim \int f_n(x)dx=\lim (2\sqrt x|_{1/n}^1)=2\Rightarrow f\in L^1(\mathbb R).$