Levi's Convergence Theorem for Series

# Levi's Convergence Theorem for Series

On the Levi's Monotone Convergence Theorems page we stated three very important convergence theorems for montoone sequence of functions (step/upper/Lebesgue-integrable) and their integrals.

We will now state another theorem known as Levi's theorem for series of Lebesgue integrable functions.

 Theorem 1 (Levi's Theorem for Series of Lebesgue Integrable Functions): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue integrable functions that are nonnegative almost everywhere on $I$ and such that $\displaystyle{\sum_{n=1}^{\infty} \int_I f_n(x) \: dx}$ converges. Then $\displaystyle{\sum_{n=1}^{\infty} f_n(x) \: dx}$ converges to a function $f \in L(I)$ almost everywhere on $I$ and $\displaystyle{\sum_{n=1}^{\infty} \int_I f_n(x) \: dx = \int_I \sum_{n=1}^{\infty} f_n(x) \: dx}$.

Like the other Levi's theorems, the proof is rather technical and so we'll omit it. We will instead look at a nice example of applying this theorem.

For example, suppose that we want to show that the $\displaystyle{\int_0^1 \ln \left ( \frac{1}{1 - x} \right ) = 1}$ as a Lebesgue integral. Consider the following sequence of functions, $(f_n(x))_{n=1}^{\infty}$:

(1)
\begin{align} \quad (f_n(x))_{n=1}^{\infty} = \left ( \frac{x^n}{n} \right )_{n=1}^{\infty} = \left ( x, \frac{x^2}{2}, \frac{x^3}{3}, ..., \frac{x^n}{n}, ... \right ) \end{align}

Notice that each $f_n$ is Lebesgue integrable on $[0, 1]$. This is because each $f_n$ is Riemann integrable on $[0, 1]$ since each $f_n$ is continuous on the closed and bounded interval $[0, 1]$, and $\displaystyle{\int_0^1 f_n(x) \: dx = \int_0^1 \frac{x^n}{n} \: dx = \frac{x^{n+1}}{n(n+1)} \biggr \rvert_0^1 = \frac{1}{n(n+1)}}$.

Also, each $f_n$ is nonnegative on $[0, 1]$ since $x \geq 0$ and $n > 1$.

Now from above, since $\displaystyle{\int_0^1 f_n(x) \: dx = \frac{1}{n(n+1)}}$ we see that the series $\displaystyle{\sum_{n=1}^{\infty} \int_I f_n(x) \: dx = \sum_{n=1}^{\infty} \frac{1}{n^2 +n}}$ actually converges by the comparison test with the convergent series $\sum_{n=1}^{\infty} \frac{1}{n^2}$.

So we can conclude that $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges to a limit function, $f$, almost everywhere on $I$ and that this limit function $f$ is Lebesgue integrable on $I$.

We claim that $\displaystyle{\sum_{n=1}^{\infty} \frac{x^n}{n}}$ converges almost everywhere to precisely $\ln \left ( \frac{1}{1 - x} \right )$ on $[0, 1]$. To see this, let $f$ denote the sum of this series. Then:

(2)
\begin{align} \quad f(x) & = \sum_{n=1}^{\infty} \frac{x^n}{n} \\ \quad f'(x) &= \sum_{n=1}^{\infty} x^{n-1} \\ \quad xf'(x) &= \sum_{n=1}^{\infty} x^n \\ \quad xf'(x) + 1 &= \sum_{n=0}^{\infty} x^n \\ \quad xf'(x) + 1 &= \frac{1}{1 - x} \\ \quad xf'(x) &= \frac{1}{1 - x} - 1 \\ \quad xf'(x) &= \frac{1}{1 - x} - \frac{1 - x}{1- x} \\ \quad xf'(x) &= \frac{x}{1 - x} \\ \quad f'(x) & = \frac{x}{x - x^2} \\ \quad f'(x) & = \frac{1}{1 - x} \\ \quad \int f'(x) \: dx &= \ln \left ( \frac{1}{1 - x} \right ) \\ \quad f(x) = \ln \left ( \frac{1}{1 - x} \right ) \end{align}

This is valid for $\mid x \mid < 1$ and in particular, for $[0, 1)$. So $\displaystyle{\sum_{n=1}^{\infty} \frac{x^n}{n}}$ converges to $f$ ALMOST EVERYWHERE on $[0, 1]$ (except at $x = 1$).

Furthermore, from the conclusion of Theorem 1 we have that:

(3)
\begin{align} \quad \int_0^1 \ln \left ( \frac{1}{1 - x} \right ) \: dx = \int_0^1 \sum_{n=1}^{\infty} \frac{x^n}{n} \: dx = \sum_{n=1}^{\infty} \int_0^1 \frac{x^n}{n} \: dx \end{align}