# Showing $f(x) = \frac{1}{\sqrt x}$ is Lebesgue integrable on $(0,1]$

While reviewing for an exam recently, I came across this question which gave me pause.

Explain why $$f(x) = \frac{1}{\sqrt x}$$ is Lebesgue integrable over $$(0,1]$$.

It is clear that $$f$$ is a decreasing, non-negative function. So, it is called Lebesgue integrable over $$(0,1]$$ if:

$$\int_{(0,1]} f \text{dm} < \infty$$

And:

$$\int_{(0,1]} f \text{dm} = \sup\{ \int_{(0,1]} s \text{dm}: s \text{ simple}, s(x) \leq f(x) \forall x \in (0,1]\}$$

The problem I’m having is that a lot of the typical useful theorems (MCT, DCT) are about non-decreasing functions.

Broadly speaking, I understand that the “problem” with this function is that $$f(0)$$ is undefined and as $$x \to 0$$, $$f(x)$$ gets very big.

I guess that the reason this function is integrable over $$(0,1]$$ is that the most “problematic” point ($$x = 0$$) is removed.

How do I go about showing this more rigorously? Which (common) theorems should I use?

• Could you, instead, think about integration of $g(x)=\dfrac{1}{\sqrt{1-x}}$ on $[0,1)?$ And then argue that $\displaystyle\int_0^1f(x)\,dm=\int_0^1g(x)\,dm?$ – Adrian Keister Dec 12 '18 at 19:56

$$\int f_{(0,1]}=\int \lim f\chi_{[1/n,1]}=\lim\int f\chi_{[1/n,1]}=\lim (2-\sqrt {2/n)}=2$$,
the first equality is obvious, the second MCT, and the third is true because the Riemann integral coincides with the Lebesgue integral on $$[1/n,1].$$