A Compute Lebesgue integral as (improper) Riemann integral

[SchroedingersLion]

Hello everyone,

in a solution to my measure theory assignment, I have seen the equation

$$
\int_{\mathbb{R}}^{} \frac {1}{|x|}\, d\lambda(x)=\infty
$$
with $\lambda$ as the 1⁻dim Lebesgue measure.

I was wondering how that integral was evaluated as we had never proven any theorem that states that Lebesgue integrals can be computed as improper Riemann integrals. This simple equation makes it look trivial, but I don't see the reasoning. Under which condition can I just rewrite the integral as $\int_{-\infty}^{\infty} \frac {1}{|x|}\, dx$?

 

[FactChecker]

I think you can easily stay within the Lebesgue integral definition. Given arbitrarily large $M \gt 0$, can you find a set of positive measure on which the function integral must be greater than $M$? Then show that the entire integral over the reals must be greater than $M$.

 

[jbunniii]

For each $n \in \mathbb Z^{\geq 1}$, let $f_n$ be the characteristic function of $(-n, -1/n) \cup (1/n, n)$. Observe that for all $x \neq 0$, $\left(\frac{1}{|x|}f_n(x)\right)_{n=1}^{\infty}$ is a nonnegative increasing sequence which converges to $1/|x|$.

Therefore by the monotone convergence theorem (second equality below), one can write
$$\begin{aligned}
\int_{\mathbb R}\frac{1}{|x|}d\lambda(x) &= \int_{\mathbb R}\frac{1}{|x|}\lim_{n \to \infty} f_n d\lambda(x) \\
&= \lim_{n \to \infty} \int_{\mathbb R}\frac{1}{|x|}f_n d\lambda(x) \\
&= \lim_{n \to \infty}\left(\int_{-n}^{-1/n}\frac{1}{|x|} d\lambda(x) + \int_{1/n}^{n}\frac{1}{|x|}d\lambda(x)\right) \\
&= \lim_{n \to \infty}\left( 2\int_{1/n}^{n}\frac{1}{x} d\lambda(x) \right)\\
&= \lim_{n \to \infty}2(\log(n) - \log(1/n)) \\
&= \lim_{n \to \infty}4\log(n) \\
&= \infty
\end{aligned}$$

 

[SchroedingersLion]

@FactChecker Nice idea, cheers!

@jbunniii Thanks for showing how to properly reduce it to a Riemann integral!

 

[Stephen Tashi]

Therefore by the monotone convergence theorem (second equality below), one can write
$\int_{\mathbb R}\frac{1}{|x|}d\lambda(x) = \int_{\mathbb R}\frac{1}{|x|}\lim_{n \to \infty} f_n d\lambda(x)$
$= \lim_{n \to \infty} \int_{\mathbb R}\frac{1}{|x|}f_n d\lambda(x)$

Often in calculus, we consider statements of the form $lim_{x \rightarrow a} f(x) = lim_{x \rightarrow a} g(x)$ to be true when both limits fail to exist - or are both exist by being "infinity". It's an interesting technical question whether the monotone covergence is stated using this convention. Does the monotone convergence theorem deal only with convergence?

 

[jbunniii]

Often in calculus, we consider statements of the form $lim_{x \rightarrow a} f(x) = lim_{x \rightarrow a} g(x)$ to be true when both limits fail to exist - or are both exist by being "infinity". It's an interesting technical question whether the monotone covergence is stated using this convention. Does the monotone convergence theorem deal only with convergence?

The monotone convergence theorem only deals with pointwise monotonically increasing sequences of nonnegative functions. Therefore the only possible type of convergence failure is divergence to $+\infty$.

For this reason among others, it's conventional in the Lebesgue context to work with the extended real numbers, in which case divergence to $+\infty$ is actually convergence to $+\infty$.

Whether or not we assume this convention, the monotone convergence theorem handles this case, in the sense that if either side diverges to $+\infty$ (or equivalently, converges in the extended reals to $+\infty$), then both do.