Proof of Levi's Theorem in Measure Theory 📂Measure Theory

Proof of Levi's Theorem in Measure Theory

Theorem ¹

If $\displaystyle \sum_{k=1}^{\infty} \int |f_{k}| dm < \infty$, then $\displaystyle \sum_{k=1}^{\infty} f_{k} (x)$ converges almost everywhere, is Lebesgue integrable, and its integral is concretely given as follows. $$ \int \sum_{k=1}^{\infty} f_{k} dm = \sum_{k=1}^{\infty} \int f_{k} dm $$

Proof

Part 1. $\displaystyle \sum_{k=1}^{\infty} f_{k} (x)$ converges almost everywhere and is Lebesgue integrable

If we define $\displaystyle \phi (x) := \sum_{k=1}^{\infty} | f_{k} (x) |$, then $\phi$ is a non-negative measurable function.

Corollary of the Monotone Convergence Theorem: Let a sequence $\left\{ f_{n} \right\}$ of measurable functions whose values are non-negative satisfy $f_{n} \nearrow f$. Then $$\int \sum_{n=1}^{\infty} f_{n} dm = \sum_{n=1}^{\infty} \int f_{n} dm$$

By the Monotone Convergence Theorem, $$ \displaystyle \int \phi dm = \sum_{n=1}^{\infty} \int | f_{k} | dm $$ and since by assumption $\displaystyle \sum_{k=1}^{\infty} \int |f_{k}| dm < \infty$, the value $\int \phi dm$ is finite, so $\phi$ is Lebesgue integrable, that is, $\phi \in \mathcal{L}^{1}$. Therefore $\phi$ is finite almost everywhere, and from the definition itself, $$ \sum_{k=1}^{\infty} \left| f_{k} \right| = \phi < \infty $$ so $\displaystyle \sum_{k=1}^{\infty} | f_{k } (x) |$ converges almost everywhere, and hence $\displaystyle \sum_{k=1}^{\infty} f_{k } (x)$ also converges almost everywhere. Meanwhile, since $$ \begin{align*} \left| \sum_{ k=1 }^{ n } f_{k} (x) \right| \le & \sum_{ k=1 }^{ n } \left| f_{k} (x) \right| \\ \le & \sum_{k=1}^{\infty} \left| f_{k} (x) \right| \\ = & \phi (x) \end{align*} $$ for all $n \in \mathbb{N}$ the inequality $\displaystyle \left| \sum_{ k=1 }^{ n } f_{k} (x) \right| \le \phi (x)$ holds almost everywhere. Finally, we define $f$, which is similar to $\phi$ but without the absolute value inside the sigma, as $$ f(x) := \sum_{k=1}^{\infty} f_{k} (x) $$ and for such an $f$, at points where it does not converge we simply set it to $f(x)$, which can guarantee that it is a Lebesgue integrable function. Accordingly, $$ \left| \sum_{ k=1 }^{ n } f_{k} \right| \le \phi \qquad , \phi \in \mathcal{L}^{1} \\ f = \lim_{n \to \infty} \sum_{k=1}^{n} f_{k} $$ so we are now ready to use the Dominated Convergence Theorem.

Part 2. The concrete integral

Dominated Convergence Theorem: For a measurable set $E \in \mathcal{M}$ and $g \in \mathcal{L}^{1} (E)$, let a sequence $\left\{ f_{n} \right\}$ of measurable functions satisfy $|f_{n}| \le g$ almost everywhere on $E$. If $\displaystyle f = \lim_{n \to \infty} f_{n}$ almost everywhere on $E$, then $$f \in \mathcal{L}^{1}(E) \\ \lim_{ n \to \infty} \int_{E} f_{n} (x) dm = \int_{E} f dm$$

Therefore the following holds. $$ \begin{align*} \int \sum_{k=1}^{\infty} f_{k} dm = & \int \lim_{n \to \infty} \sum_{k=1}^{n} f_{k} dm \\ =& \int f dm \\ =& \lim_{n \to \infty} \int \sum_{k=1}^{n} f_{k} dm \\ =& \lim_{n \to \infty} \sum_{k=1}^{n} \int f_{k} dm \\ =& \sum_{k=1}^{\infty} \int f_{k} dm \end{align*} $$

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Explanation

This theorem, proved by the Italian mathematician beppo Levi, makes the integration of a sequence of functions astonishingly easy.

Example

Evaluate $\displaystyle \int_{0}^{1} \left( {{ \log x } \over {1-x}} \right)^2 dx$.

Solution

If we define $f_{n} (x) : = n x^{n-1} ( \log x )^2$, then since $\displaystyle \sum_{n=1}^{\infty} n x^{n-1} = {{1} \over {(1-x)^2}}$, $$ \sum_{n=1}^{\infty} f_{n} (x) = \left( {{ \log x } \over {1-x}} \right)^2 $$ holds. By Levi’s Theorem, $$\int_{0}^{1} f (x) dx = \sum_{n=1}^{\infty} \int_{0}^{1} f_{n} (x) dx$$ and by integration by parts, $$\int_{0}^{1} f_{n} (x) dx = \int_{0}^{1} n x^{n-1} ( \log x )^2 dx = {{2} \over {n^2}}$$ Therefore we obtain the following. $$\displaystyle \int_{0}^{1} f(x) dx = \sum_{n=1}^{\infty} {{2} \over {n^2}} = {{\pi^2} \over {3}}$$

Commentary

The key to solving the problem is ultimately to find an $f_{n}$ satisfying $ \begin{cases} \displaystyle \sum_{n=1}^{\infty} f_{n} (x) = f(x) \\ \displaystyle \int_{E} f(x) dx = \sum_{n=1}^{\infty} \int_{E} f_{n} (x) dx \end{cases}$. It is true that this process is by no means easy, but it will surely be far easier than solving it by finding an antiderivative as in Riemann integration.