Beppo Levi's Lemma for Nonnegative Increasing Measurable Functions

# Beppo Levi's Lemma for Nonnegative Increasing Measurable Functions

Lemma 1: Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be an extended nonnegative measurable function defined on a measurable set $E$ such that $\displaystyle{\int_E f(x) \: d \mu < \infty}$. Then:a) The set $\{ x \in E : f(x) = \infty \}$ has finite measure.b) The set $\{ x \in E : f(x) > 0 \}$ is $\sigma$-finite. |

**Proof of a)**Let:

\begin{align} \quad E^* = \{ x \in E : f(x) = \infty \} \end{align}

- That is, $E^*$ is the set of all elements in the domain of $f$ that are mapped to $\infty$. Then the simple function $\varphi$ on $E$ defined by $\varphi(x) = 1$ if $x \in E^*$ and $\varphi(x) = 0$ if $x \in E \setminus E^*$ is such that $0 \leq \varphi(x) \leq f(x)$ on $E$. So:

\begin{align} \quad \mu(E^*) = \int_{E^*} \varphi(x) \: d\mu = \int_E \varphi(x) \: d \mu \leq \int_E f(x) \: d \mu < \infty \end{align}

- So $\mu(E^*)$ has finite measure. $\blacksquare$

**Proof of b)**For each $n \in \mathbb{N}$ let:

\begin{align} \quad E_n = \left \{ x \in E : f(x) > \frac{1}{n} \right \} \end{align}

- Then clearly $\displaystyle{\{ x \in E : f(x) > 0 \} = \bigcup_{n=1}^{\infty} E_n}$. Moreover, for each $n \in \mathbb{N}$ we have that:

\begin{align} \quad \frac{1}{n} \cdot \mu (E_n) = \int_{E_n} \frac{1}{n} \: d \mu \leq \int_{E_n} f(x) \: d \mu \leq \int_E f(x) \: d \mu < \infty \end{align}

- Therefore $\mu (E_n) < \infty$. So $\{ x \in E : f(x) > 0 \}$ is a countable union of measurable sets that have finite measure. So $\{ x \in E : f(x) > 0 \}$ is $\sigma$-finite. $\blacksquare$

Beppo Levi's Lemma for Nonnegative Increasing Measurable Functions: Let $(X, \mathcal A, \mu)$ be a measure space and let $(f_n(x))_{n=1}^{\infty}$ be a pointwise increasing sequence of nonnegative measurable functions defined on a measurable set $E$. If the numerical sequences $\left ( \int_E f_n(x) \: d \mu \right )_{n=1}^{\infty}$ is bounded, then:a) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to a measurable function $f$ on $E$.b) $\displaystyle{\lim_{n \to \infty} \int_E f_n(x) \: d \mu = \int_E f(x) \: d \mu}$.c) The set $\{ x \in E : f(x) = \infty \}$ has finite measure. |

**Proof of a):**For each $x \in E$ let:

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

- Then $f$ is a measurable function as it is a pointwise limit of measurable functions, and by definition, $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$. $\blacksquare$

**Proof of b)**Now $(f_n(x))_{n=1}^{\infty}$ is a nonnegative sequence of measurable functions such that $f_1(x) \leq f_2(x) \leq ... \leq f_n(x) \leq ... \leq f(x)$ that converges pointwise to $f(x)$. So by The Monotone Convergence Theorem for Nonnegative Measurable Functions we have that:

\begin{align} \quad \lim_{n \to \infty} \int_E f_n(x) \: d \mu = \int_E f(x) \: d \mu \end{align}

**Proof of c)**The numerical sequence $\left ( \int_E f_n(x) \: d \mu \right )_{n=1}^{\infty}$ is bounded. So the limit, $\displaystyle{\int_E f(x) \: d \mu < \infty}$. By Lemma 1 we have that $\{ x \in E : f(x) = \infty \}$ has finite measure. $\blacksquare$