Lebesgue's Dominated Convergence Theorem for Series

Recall from the Lebesgue's Dominated Convergence Theorem page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions on $I$ that converge almost everywhere to a limit function $f$ on $I$ and if there exists a Lebesgue integrable function $g(x)$ on $I$ such that $\mid f_n(x) \mid \leq g(x)$ almost everywhere on $I$ and for all $n \in \mathbb{N}$ then $f$ is Lebesgue integrable on $I$ and:

(1)

\begin{align} \quad \int_I f(x) \: dx = \int_I \lim_{n \to \infty} f_n(x) \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx \end{align}

We will now look at a related theorem known as Lebesgue's dominated convergence theorem for series.

Theorem 1 (Lebesgue's Dominated Convergence Theorem for Series): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue integrable functions on $I$ such that each $f_n$ is nonnegative on $I$ and such that $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges almost everywhere on $I$ to a limit function $f$ and suppose that there exists a nonnegative Lebesgue integrable function $g$ on $I$ such that $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$. Then $f$ is Lebesgue integrable on $I$ and $\displaystyle{\int_I f(x) \: dx = \int_I \sum_{n=1}^{\infty} f_n(x) \: dx = \sum_{n=1}^{\infty} \int_I f_n(x) \: dx}$.

• Proof: Define a new sequence of functions $(g_n(x))_{n=1}^{\infty}$ for each $n \in \mathbb{N}$ by:

(2)

\begin{align} \quad g_n(x) = \sum_{k=1}^{n} f_k(x) \end{align}

• From the Linearity of Lebesgue Integrals page we know that the sum of Lebesgue integrable functions is Lebesgue integrable, and so each $g_n$ is a Lebesgue integrable function and moreover, $(g_n(x))_{n=1}^{\infty}$ converges to $f$ since:

(3)

\begin{align} \quad \lim_{n \to \infty} g_n(x) = \lim_{n \to \infty} \sum_{k=1}^{n} f_k(x) \: dx = \sum_{k=1}^{\infty} f_k(x) = \sum_{n=1}^{\infty} f_n(x) = f(x) \end{align}

• Moreover, since each $g_n(x)$ is nonnegative on $I$ we see that for each $n \in \mathbb{N}$ that $g_n(x) = \mid g_n(x) \mid \leq f(x)$ almost everywhere on $I$. Furthermore, since $f(x)$ is bounded above by a Lebesgue integrable function $g(x)$ almost everywhere on $I$ we see that $\mid g_n(x) \mid \leq g(x)$ almost everywhere on $I$. So by Lebesgue's dominated convergence theorem we have that the limit function, $f$ is Lebesgue integrable on $I$ and moreover:

(4)

\begin{align} \quad \int_I f(x) \: dx = \int_I \lim_{n \to \infty} g_n(x) \: dx = \int_I \lim_{n \to \infty} \sum_{k=1}^{n} f_k(x) \: dx & = \int_I \sum_{n=1}^{\infty} f_n(x) \: dx \\ \quad &= \lim_{n \to \infty} \int_I g_n(x) \: dx \\ \quad &= \lim_{n \to \infty} \int_I \sum_{k=1}^{n} f_k(x) \: dx \\ \quad &= \lim_{n \to \infty} \int_I [f_1(x) + f_2(x) + ... + f_n(x)] \: dx \\ \quad &= \lim_{n \to \infty} \left ( \int_I f_1(x) \: dx + \int_I f_2(x) \: dx + ... + \int_I f_n(x) \: dx \right ) \\ \quad &= \lim_{n \to \infty} \sum_{k=1}^{n} \int_I f_k(x) \: dx \\ \quad &= \sum_{n=1}^{\infty} \int_I f_n(x) \: dx \end{align}

• Therefore we see that:

(5)

\begin{align} \quad \int_I f(x) \: dx = \int_I \sum_{n=1}^{\infty} f_n(x) \: dx = \sum_{n=1}^{\infty} \int_I f_n(x) \: dx \quad \blacksquare \end{align}