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:
\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:
\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.
- From Fatou's Lemma for Nonnegative Measurable Functions we have that:
\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:
\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:
\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:
\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}