16.2 Integration of Non-Negative Functions

Definition 16.2.1. Let $(X, \cm)$ be a measure space, then

\[\mathcal{L}^{+}(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}\]

is the space of non-negative $\real$-valued measurable functions on $(X, \cm)$.

Definition 16.2.2 (Integral of Non-Negative Function). Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^{+}(X, \cm)$, then

\[\int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}\]

is the Lebesgue integral of $f$.

Lemma 16.2.3. Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^{+}(X, \cm)$, then

\[\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}\]

Proof. Let $\phi \in \Sigma^{+}(X, \cm)$ with $\phi \le f$, then for any $\alpha \in (0, 1)$, $\alpha \phi|_{\bracs{f > 0}}< f|_{\bracs{f > 0}}$. Since

\[\int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu\]

the two sides are equal.$\square$

Theorem 16.2.4 (Monotone Convergence Theorem, [Theorem 2.14, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n}\subset \mathcal{L}^{+}(X, \cm)$, and $f \in \mathcal{L}^{+}(X, \cm)$ such that $f_{n} \upto f$ pointwise, then

\[\limv{n}\int f_{n} d\mu = \int f d\mu\]

Proof. By Definition 16.2.2, $\int f_{n} d\mu \le \int f d\mu$ for each $n \in \natp$.

Let $\phi \in \Sigma^{+}(X, \cm)$ with $\phi|_{\bracs{f > 0}}< f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_{n} \upto f$, $\bracs{f_n \ge \phi}\upto X$. In which case, since $A \mapsto \int \one_{A} \phi d\mu$ is a measure (Proposition 16.1.3),

\[\limv{n}\int f_{n} d\mu \ge \limv{n}\int \one_{\bracs{f_n \ge \phi}}\cdot \phi d\mu = \int \phi d\mu\]

by continuity from below (Proposition 14.1.5). Therefore $\limv{n}\int f_{n} d\mu \ge \int f$ by Lemma 16.2.3.$\square$

Lemma 16.2.5 (Fatou, [Lemma 2.18, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n}\subset \mathcal{L}^{+}(X, \cm)$, then

\[\int \liminf_{n \to \infty}f_{n} d\mu \le \liminf_{n \to \infty}\int f_{n}d\mu\]

Proof. For each $n \in \natp$, $\inf_{k \ge n}f_{k} \le f_{n}$. By the monotone convergence theorem,

\[\int \liminf_{n \to \infty}f_{n} d\mu = \limv{n}\int \inf_{k \ge n}f_{k} d\mu \le \liminf_{n \to \infty}\int f_{n} d\mu\]

$\square$

Lemma 16.2.6. Let $(X, \cm)$ be a measurable space and $f \in \mathcal{L}^{+}(X, \cm)$, then there exists $\seq{f_n}\subset \Sigma^{+}(X, \cm)$ such that $f_{n} \upto f$ pointwise.

Proof. By Proposition 15.5.6, there exists $\seq{g_n}\subset \Sigma^{+}(X, \cm)$ such that $0 \le g_{n} \le f$ for each $n \in \natp$, and $g_{n} \to f$ pointwise. For each $n \in \natp$, let $f_{n} = \max_{1 \le k \le n}g_{k}$, then $f_{n} \in \Sigma^{+}(X, \cm)$, $0 \le f_{n} \le f$, and $f_{n} \upto f$ pointwise.$\square$

Proposition 16.2.7. Let $(X, \cm, \mu)$ be a measure space, then

For any $f, g \in \mathcal{L}^{+}(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
For any $\seq{f_n}\subset \mathcal{L}^{+}(X, \cm)$, $\sum_{n \in \natp}\int f_{n}d\mu = \int \sum_{n \in \natp}f_{n} d\mu$.
For any $f, g \in \mathcal{L}^{+}(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
For any $f \in \mathcal{L}^{+}(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
For any $f \in \mathcal{L}^{+}(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
For any $f \in \mathcal{L}^{+}(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.

Proof. (1): By Lemma 16.2.6, there exists $\seq{f_n}, \seq{g_n}\subset \Sigma^{+}(X, \cm)$ with $0 \le f_{n} \le f$ and $0 \le g_{n} \le g$ for each $n \in \natp$, $f_{n} \upto f$, and $g_{n} \upto g$. By Proposition 16.1.3 and the Monotone Convergence Theorem,

\begin{align*}\int \alpha f + g d\mu = \limv{n}\int \alpha f_{n} + g_{n} d\mu = \alpha\limv{n}\int f_{n} d\mu + \limv{n}\int g_{n} d\mu \\&= \alpha \int f d\mu + \int g d\mu\end{align*}

(2): By (1) and the Monotone Convergence Theorem,

\[\int \sum_{n \in \natp}f_{n} d\mu = \int \limv{N}\sum_{n = 1}^{N} f_{n} d\mu = \limv{N}\sum_{n = 1}^{N} \int f_{n}d\mu = \sum_{n \in \natp}\int f_{n} d\mu\]

(3): By Definition 16.2.2.

(4): Since $\int f d\mu \ge \int \infty \cdot \one_{\bracs{f = \infty}}d\mu = \infty \cdot \mu(\bracs{f = \infty})$, $\int f d\mu < \infty$ implies that $\mu(\bracs{f = \infty}) = 0$.

(5): For each $\eps > 0$,

\[\int f d\mu \ge \int \eps \cdot \one_{\bracs{f \ge \eps}}= \eps \cdot \mu(\bracs{f \ge \eps})\]

so $\int f d\mu < \infty$ implies that $\mu(\bracs{f \ge \eps}) < \infty$. Since $\bracs{f > 0}= \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$, $\bracs{f > 0}$ is $\sigma$-finite.

(6): As in the proof of (5), $\mu(\bracs{f \ge \eps}) = 0$ for all $\eps > 0$. Thus $\bracs{f > 0}= \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$ is a $\mu$-null set, so $f = 0$ $\mu$-almost everywhere.$\square$

Jerry's Digital Garden

16.2 Integration of Non-Negative Functions

Direct References

Jerry's Digital Garden

Direct References