Lebesgue integrability of $x^{-\frac{1}{2}}$

Question

Is the function $x^{-\frac{1}{2}}$ is Lebesgue integrable over $\mathbb R$ or on some subset of $\mathbb R$?

Answer 1

Yes, of course: it is integrable over any interval $[a,b]$ with $0<a<b<+\infty$, since it is continuous on such an interval.

It is not integrable on $(a,\infty)$, $a<\infty$, since $$ \lim_{b \to \infty} \int_a^b \frac{dx}{\sqrt{x}} = +\infty. $$

Answer 2

There is a potential problem at $x=0$ since the function diverges there. Also, there is a problem for $x<0$ since you seem to not want to use complex numbers.

So the function is integrable on any finite subset not including zero or negative numbers.

However, it doesn't matter what value the function takes at zero since $\{0\}$ is a set of measure $0$. Define $g(x)=f(x)$ for all $x$ except $g(0)=0$. In fact $\int_A g d\lambda=\int_A f d\lambda$ for any measureable set $A$ that doesn't contain zero or negative numbers. So we have found a function that is Lebesgue integrable on $[0,b]$, $\forall b>0$ which is identical to $f$ everywhere except on a set of measure zero.

In particular, $$\lim_{a\rightarrow 0+} \int_a^b \frac{1}{\sqrt{x}} dx=\lim_{a\rightarrow 0+} 2(\sqrt{b}-\sqrt{a})=2\sqrt{b}.$$ So the function is improperly integrable in $[0,b]$. I think it is usually considered Lebesgue integrable as well when it is improperly Riemann integrable.

Edit: as noted in comments below, improper Riemann integrability does not necessarily imply Lebesgue integability as the latter requires both $|f_+|$ and $|f_-|$ to have finite integrals. The integral is said to exist if at least one is finite. This is according to conventions in Measure Theory by Cohn (p.65 1st edition).

Answer 3

If you consider the function which is $x^{-1/2}$ except at zero, where it is arbitrarily defined, this function is Lebesgue integrable on any bounded subset of $\mathbb{R}_+$, even if it contains zero. Here's a proof:

1. Note that the proper Riemann integral and Lebesgue integral always agree, so if $0<a<b$ then $\int_a^b x^{-1/2} dx$ is the same in the Riemann and Lebesgue sense.
2. Justify that $\int_{1/n}^b x^{-1/2} dx \to \int_0^b x^{-1/2} dx$ as $n \to \infty$ in the Lebesgue sense. The latter will always be defined (after choosing a value for $x^{-1/2}$ at zero) since this function is measurable and nonnegative. The only concern is that it might be infinite. This can be done with monotone convergence, since $\int_{1/n}^b x^{-1/2} dx = \int_0^b x^{-1/2} \chi_{[1/n,b]}(x) dx$.
3. Use the fact that $\int_{1/n}^b x^{-1/2} dx = 2 b^{1/2} - 2 n^{-1/2}$ from Riemann integration, then send $n \to \infty$.