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:

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


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

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?

  • $\begingroup$ Could you, instead, think about integration of $g(x)=\dfrac{1}{\sqrt{1-x}}$ on $[0,1)?$ And then argue that $\displaystyle\int_0^1f(x)\,dm=\int_0^1g(x)\,dm?$ $\endgroup$ – 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 $[1/n,1].$

As you point out, one strategy for doing these problems is to "back off" the bad point and use MCT or DCT.


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