[proofplan] We reduce to the [Monotone Convergence Theorem](/theorems/509) by constructing the monotone minorant $g_n = \inf_{m \geq n} f_m$, which converges pointwise to $\liminf f_n$ and satisfies $g_n \leq f_n$. The MCT evaluates $\int \liminf f_n = \lim \int g_n$, and the domination $g_n \leq f_n$ converts this into the desired inequality $\lim \int g_n \leq \liminf \int f_n$. [/proofplan] [step:Construct the monotone minorant $g_n = \inf_{m \geq n} f_m$ and verify its properties] Define $g_n: E \to [0, \infty]$ by $g_n = \inf_{m \geq n} f_m$ for each $n \in \mathbb{N}$. [claim:Minorant Properties] The sequence $(g_n)$ satisfies: (i) each $g_n$ is measurable; (ii) $0 \leq g_n \leq g_{n+1}$ for all $n$; (iii) $g_n(x) \to \liminf_{m \to \infty} f_m(x)$ for every $x \in E$; (iv) $g_n \leq f_n$ pointwise. [/claim] [proof] For (i): $g_n = \inf_{m \geq n} f_m$ is an infimum of countably many measurable functions. For any $y \in \mathbb{R}$, $\{g_n > y\} = \bigcap_{m \geq n} \{f_m > y\}$, which is a countable intersection of measurable sets and hence measurable. So $g_n$ is measurable. For (ii): since $\{f_m : m \geq n+1\} \subseteq \{f_m : m \geq n\}$, taking the infimum over the smaller collection yields a value at least as large: $g_{n+1} = \inf_{m \geq n+1} f_m \geq \inf_{m \geq n} f_m = g_n$. Non-negativity follows from $f_m \geq 0$. For (iii): by definition, $\liminf_{m \to \infty} f_m(x) = \sup_{n \geq 1} \inf_{m \geq n} f_m(x) = \sup_{n \geq 1} g_n(x) = \lim_{n \to \infty} g_n(x)$, where the last equality holds because $(g_n(x))$ is non-decreasing. For (iv): $g_n = \inf_{m \geq n} f_m \leq f_n$, since $f_n$ is one of the functions in the infimum. [/proof] [/step] [step:Apply the Monotone Convergence Theorem to $(g_n)$] By (i)--(iii), $(g_n)$ is a non-decreasing sequence of non-negative measurable functions converging pointwise to $\liminf_{n} f_n$. The [Monotone Convergence Theorem](/theorems/509) gives \begin{align*} \int_E \liminf_{n \to \infty} f_n \, d\mu = \lim_{n \to \infty} \int_E g_n \, d\mu. \end{align*} [/step] [step:Conclude via the domination $g_n \leq f_n$] By (iv) and monotonicity of the integral, $\int_E g_n \, d\mu \leq \int_E f_n \, d\mu$ for each $n$. Since $(g_n)$ is non-decreasing, the sequence $(\int_E g_n \, d\mu)$ is also non-decreasing, so its limit equals its liminf. Therefore \begin{align*} \int_E \liminf_{n \to \infty} f_n \, d\mu = \lim_{n \to \infty} \int_E g_n \, d\mu = \liminf_{n \to \infty} \int_E g_n \, d\mu \leq \liminf_{n \to \infty} \int_E f_n \, d\mu, \end{align*} where the inequality uses $\int g_n \, d\mu \leq \int f_n \, d\mu$ and the general fact that $a_n \leq b_n$ implies $\liminf a_n \leq \liminf b_n$. [/step]