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Proposition

$f(x) = 1/x$ is in $L^2\left([1, +\infty)\right)$ but not in $L^1\left([1, +\infty)\right)$.

Discussion

So my issue here is that I don't know how to use infinity in Lebesgue integration.

It is intuitive (I think) that evaluation of the improper Riemann integrals

\begin{align} \int_1^\infty \left|f(x)\right| &= \int_1^\infty \frac{1}{x} = \lim_{c \to \infty} \ln c = + \infty \\ \\ \int_1^\infty \left|f(x)\right|^2 &= \int_1^\infty \frac{1}{x^2} = 1 - \lim_{c \to \infty} \frac{1}{c} = 1 \end{align}

would imply our proposition, but I've only seen $L^p$-spaces defined in the sense of Lebesgue integrals. So when I get to these steps:

\begin{align} \int_{[1, \infty)} \left|f(x)\right| &= \int_{[1, \infty)} \frac{1}{x} = \cdots \\ \\ \int_{[1, \infty)} \left|f(x)\right|^2 &= \int_{[1, \infty)} \frac{1}{x^2} = \cdots \end{align}

I'm not sure how to proceed. I'm guessing we need an argument for switching between the two types of integration, which I've read up on a little bit, but am not sure how to apply here in the improper case.

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  • $\begingroup$ Try breaking the domain into a disjoint, countable sum of intervals $[n,n+1]$, integrating over each interval, and then apply the monotone convergence theorem to conclude that the the integral over the countable union is indeed the infinite sum of the integrals over each union. $\endgroup$ Mar 3, 2019 at 20:52

2 Answers 2

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Absolutely convergent improper Riemann integrals are also Lebesgue integrals. For the other direction, use the monotonicity property $$ \int_{A}|f|\leq\int_{B}|f| $$ for measurable sets $A\subset B$.

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There seems to be some consternation relating to the definition of improper Lebesgue integrals. See this related post for some discussion. In your situation we can avoid this difficulty by using the Monotone Convergence Theorem. Clearly, if we define $f_n:=\mathbb 1_{[1,n]}f$, for every $n\in\mathbb N$ we have that $f_{n}\leq f_{n+1}$, and the sequence $(f_n)$ converges pointwise to $f$. The MCT thus gives that $f$ is Lebesgue measurable, and $$\int_{[1,\infty)}fdx=\lim_{n\to \infty}\int_{[1,\infty)}f_ndx=\lim_{n\to\infty}\int_{[1,n]}fdx.$$ Thus in your case the integral evaluation boils down to evaluation of the improper Riemann integral, without ever having to talk about improper Lebesgue integration. Note that technically Lebesgue integrals aren't signed, so you shouldn't really have upper and lower limits of the integral.

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