The Monotone Conv. Theo. for Nonnegative Measurable Functions

# The Monotone Convergence Theorem for Nonnegative Measurable Functions

## The Monotone Convergence Theorem for Nonnegative Measurable Functions

Recall from the Lebesgue's Monotone Convergence Theorem page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$ such that:

• 1) $0 \leq f_1(x) \leq f_2(x) \leq ... \leq f_n(x) \leq ... \leq f(x)$ on $E$.
• 2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$.

Then $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$.

We now state an analogous monotone convergence theorem for general complete measure spaces.

 Theorem 1 (The Monotone Convergence Theorem for General Measure Spaces): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative measurable functions defined on a measurable set $E$ such that: 1) $0 \leq f_1(x) \leq f_2(x) \leq ... \leq f_n(x) \leq ... \leq f(x)$ on $E$. 2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$. Then $\displaystyle{\lim_{n \to \infty} \int_E f_n(x) \: d \mu = \int_E f(x) \: d \mu}$.
• Proof: For each $n \in \mathbb{N}$, since $0 \leq f_n(x) \leq f(x)$ we have that:
(1)
\begin{align} \quad \int_E f_n(x) \: d \mu \leq \int_E f(x) \: d \mu \end{align}
• Since this holds for all $n \in \mathbb{N}$ we have that:
(2)
\begin{align} \quad \lim_{n \to \infty} \int_E f_n(x) \: d \mu \leq \int_E f(x) \: d \mu \quad (*) \end{align}
• We also note that $\displaystyle{\left ( \int_E f_n(x) \: d \mu \right )_{n=1}^{\infty}}$ is an increasing sequence of nonnegative real numbers that is bounded above, and hence converges, so $\displaystyle{\lim_{n \to \infty} \int_E f_n(x) \: d \mu}$ exists.
(3)
\begin{align} \quad \int_E f(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu = \lim_{n \to \infty} \int_E f_n(x) \: d \mu \quad (**) \end{align}
• From $(*)$ and $(**)$ we conclude that:
(4)
\begin{align} \quad \lim_{n \to \infty} \int_E f_n(x) \: d \mu = \int_E f(x) \: d \mu \end{align}

## The Monotone Convergence Theorem for Series of Nonnegative Measurable Functions

Recall from the Lebesgue's Monotone Convergence Theorem for Series page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$ such that:

• 1) $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges on $E$.

Then $\displaystyle{\sum_{n=1}^{\infty} \int_E f_n(x) \: dx = \int_E \sum_{n=1}^{\infty} f_n(x) \: dx}$. ||

Once again, we can state and prove an analogous theorem for general measure spaces.

 Theorem 2 (The Monotone Convergence Theorem for Series of Nonnegative Measurable Functions): Let $(X, \mathcal A, \mu)$ be a measure space. If $(f_n(x))_{n=1}^{\infty}$ is a sequence of nonnegative measurable functions defined on a measurable set $E$ such that: 1) $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges on $E$. Then $\displaystyle{\sum_{n=1}^{\infty} \int_E f_n(x) \: d \mu = \int_E \sum_{n=1}^{\infty} f_n(x) \: d\mu}$.
• Proof: Consider the sequence of partial sums $(u_n(x))_{n=1}^{\infty}$ defined for each $n \in \mathbb{N}$ by:
(5)
\begin{align} \quad u_n(x) = \sum_{k=1}^{n} f_n(x) \end{align}
• Then $(u_n(x))_{n=1}^{\infty}$ is a sequence of nonnegative measurable functions defined on a measurable set $E$. Furthermore, $0 \leq u_1(x) \leq u_2(x) \leq ... \leq u_n(x) \leq ...$, and $(u_n(x))_{n=1}^{\infty}$ converges to $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ on $E$. So by the Monotone convergence theorem we have that:
(6)
\begin{align} \quad \lim_{n \to \infty} \int_E u_n(x) \: d \mu &= \int_E \sum_{n=1}^{\infty} f_n(x) \: d \mu \\ \quad \lim_{n \to \infty} \left [ \int_E f_1(x) \: d \mu + \int_E f_2(x) \: d \mu + ... + \int_E f_n(x) \: d \mu \right ] &= \int_E \sum_{n=1}^{\infty} f_n(x) \: d \mu \\ \quad \lim_{n \to \infty} \sum_{k=1}^{n} \int_E f_k(x) \: d \mu &= \int_E \sum_{n=1}^{\infty} f_n(x) \: d \mu \\ \quad \sum_{n=1}^{\infty} \int_E f_n(x) \: d \mu &= \int_E \sum_{n=1}^{\infty} f_n(x) \: d \mu \quad \blacksquare \end{align}