[proofplan] We show both $\int f \, d\mu \leq \lim \int f_n \, d\mu$ and $\lim \int f_n \, d\mu \leq \int f \, d\mu$. The first inequality is immediate from monotonicity. For the second, we approximate $f$ from below by simple functions $\phi \leq f$, use a parameter $\alpha \in (0,1)$ to define the sets $B_n = \{f_n \geq \alpha\phi\}$ that exhaust $E$, bound $\int f_n$ from below via the restricted integral over $B_n$, and pass to the limit using continuity of measure. [/proofplan] [step:Obtain the easy inequality $\lim \int f_n \leq \int f$ from monotonicity] Since $f_n \leq f$ pointwise for all $n$, monotonicity of the integral gives $\int_E f_n \, d\mu \leq \int_E f \, d\mu$ for each $n$. Taking the limit, $\lim_{n \to \infty} \int_E f_n \, d\mu \leq \int_E f \, d\mu$. The limit exists because the sequence of integrals is non-decreasing. [/step] [step:Reduce to showing $\int \phi \, d\mu \leq \lim \int f_n \, d\mu$ for every simple $\phi \leq f$] It suffices to show that for every simple function $\phi$ with $0 \leq \phi \leq f$, \begin{align*} \int_E \phi \, d\mu \leq \lim_{n \to \infty} \int_E f_n \, d\mu, \end{align*} since taking the supremum over all such $\phi$ recovers $\int_E f \, d\mu$ by definition of the Lebesgue integral. [/step] [step:Construct the exhausting sets $B_n = \{f_n \geq \alpha\phi\}$ and verify they increase to $E$] Fix a simple function $\phi = \sum_{j=1}^k a_j \mathbb{1}_{A_j}$ with $0 \leq \phi \leq f$, and fix $\alpha \in (0,1)$. Define the sets \begin{align*} B_n = \{ x \in E : f_n(x) \geq \alpha \phi(x) \}. \end{align*} [claim:B N Increases To E] The sequence $(B_n)$ is increasing and $\bigcup_{n=1}^\infty B_n = E$. [/claim] [proof] Since $f_n \leq f_{n+1}$, if $x \in B_n$ then $f_{n+1}(x) \geq f_n(x) \geq \alpha\phi(x)$, so $x \in B_{n+1}$. For any $x \in E$, either $\phi(x) = 0$ (in which case $x \in B_1$) or $\phi(x) > 0$, and then $f(x) \geq \phi(x) > \alpha\phi(x)$, so since $f_n(x) \uparrow f(x)$, eventually $f_n(x) \geq \alpha\phi(x)$, placing $x \in B_n$ for large $n$. [/proof] [/step] [step:Bound $\int f_n$ from below via the restricted integral over $B_n$] Since $f_n \geq \alpha\phi$ on $B_n$, \begin{align*} \int_E f_n \, d\mu \geq \int_{B_n} f_n \, d\mu \geq \alpha \int_{B_n} \phi \, d\mu = \alpha \sum_{j=1}^k a_j \mu(A_j \cap B_n). \end{align*} As $n \to \infty$, $A_j \cap B_n \uparrow A_j$ for each $j$, so by continuity of measure from below, $\mu(A_j \cap B_n) \to \mu(A_j)$. Therefore \begin{align*} \lim_{n \to \infty} \int_E f_n \, d\mu \geq \alpha \sum_{j=1}^k a_j \mu(A_j) = \alpha \int_E \phi \, d\mu. \end{align*} [/step] [step:Send $\alpha \uparrow 1$ and take the supremum over $\phi$ to conclude] Taking $\alpha \uparrow 1$, we get $\lim_{n} \int_E f_n \, d\mu \geq \int_E \phi \, d\mu$. Taking the supremum over all simple $\phi \leq f$ gives $\lim_n \int_E f_n \, d\mu \geq \int_E f \, d\mu$. Combined with the easy inequality from the first step, equality holds. [/step]