Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], and $\seq{f_j} \in L^+$ be a [[Sequence|sequence]] of $\cm$-[[Measurable Function|measurable functions]]. If $f_j \le f_{j+1}$ and $\limv{j}f_j = f$ [[Pointwise Convergence|pointwise]], then the [[Integral|integral]] can be interchanged with the [[Limit|limit]]: $ \limv{j}\int f_j d\mu = \int \limv{j}f_j d\mu $ *Proof*. # Uniform Convergence First off, if $X$ is finite, $f$ is finite, and $f_n \to f$ [[Uniform Convergence|uniformly]], then we can simply take some $\varepsilon > 0$ and $\exists N \in \nat:$ $ \int f_n d\mu + \int \frac{\varepsilon}{\mu(X)} d\mu \ge \int f d\mu \quad \forall n \ge N $ which gives $ \int f_n d\mu \ge \int f d\mu - \varepsilon $ and makes the convergence fairly straightforward. # Dropping Uniform Convergence The first hurdle comes from the lack of uniform convergence. If it's just a sequence of sequences, then we can't set an $\varepsilon$, take the max $N$ and expect everyone to line-up. There is no large enough $N$ for every sequence to stay within the bound, and no way of tracking the "progress" of the convergence. ![[monotone_convergence_measure.png]] Since we have a measure space at our disposal, measuring the chunk of space where $f - f_n \le \varepsilon$ allows us to check the "progress" of the convergence. Let $g(x) = f - f_n$, then $g$ is measurable and $g^{-1}([0, \varepsilon])$ is a measurable set. Take $ E_n = \bracs{x \in X: f_n(x) + \varepsilon \ge f(x)} = g^{-1}([0, \varepsilon]) $ then $ \int \braks{f_n + \frac{\varepsilon}{\mu(X)}} \ge \int_{E_n} \braks{f_n + \frac{\varepsilon}{\mu(X)}} \ge \int_{E_n}f $ Since $f_n \to f$, this "progress" must increase and must reach 100%. $ \forall x \in X, \exists N \in \nat: f_n + \varepsilon \ge f \Rightarrow x \in E_n \quad \forall n \ge N $ which gives $\bigcup_{n \in \nat}E_n = X$. The definite integral is also a measure, allowing us to take advantage of continuity from below by taking the limit: $ \limv{n}\int f_n + \int \frac{\varepsilon}{\mu(X)} \ge \int f \quad \limv{n}\int f_n \ge \int f - \varepsilon $ The $\varepsilon$ controls how close to $\int f$ the limit is, and the "progress bar" controls how close to the limit $\int f_n$ is. Together, they push the limit right to $\int f$. # Dropping Finite Measure Space Let's first make sure that $\int \alpha f = \alpha \int f$ ($\alpha > 0$). For any simple function $\phi \le f$, $\alpha \phi \le \alpha f$, and for any simple function $\phi \le \alpha f$, $\phi/\alpha \le f$. This means that $ \begin{align*} \alpha \int f d\mu &= \alpha \sup \bracs{\int \phi d\mu: \phi \le f} \\ &= \sup \bracs{\alpha \int \phi d\mu: \phi \le f} \\ &= \sup \bracs{\int \alpha \phi d\mu: \phi \le f} \\ &= \int \alpha f d\mu \end{align*} $ we can push the $\alpha$ all the way in. Using the $\alpha$ to modify the function and the integral allows us to remove the possibility of $\int f - \infty$ shenanigans, as we can simply take the $\alpha$ out of the integral. Using the same reasoning as before, let $\alpha \in (0, 1)$ and $ E_n = \bracs{x \in X: f_n(x) \ge \alpha f(x)} $ then $ \forall x \in X, \exists N \in \nat: f(x) - f_n(x) \le (1 - \alpha)f(x) $ where $f(x) > 0 \Rightarrow (1 - \alpha)f(x) > 0$ and we can extract the $N$ from the convergence, or $f(x) = 0$ and $f_n(x) = 0 \ge \alpha f(x)$ anyways. We still get to use $\bigcup_{n \in \nat}E_n$. Set up the same inequalities, $ \int f_n \ge \int_{E_n}f_n \ge \int_{E_n}\alpha f = \alpha\int_{E_n}f $ take the limits and obtain, $ \limv{n}\int f_n \ge \alpha \int f $ Since this result holds for all $\alpha < 1$, we have $\limv{n}\int f_n \ge \int f$. Here $\alpha$ does a similar job as the $\varepsilon$ from before, but it controls the limit better by virtue of being outside of the integral. When $\int f$ is finite, $\int \varepsilon$ doesn't have to be finite, but $(1 - \alpha)\int f$ is. When $\int f$ is infinite, using $\alpha$ guarantees the existence of some $N$ where $f_n: n \ge N$ onwards also have an infinite integral. # Dropping Finite Function Lastly, we can exploit the definition $ \int f d \mu = \sup \bracs{\int \phi d\mu: \phi \le f} $ and the fact that simple functions are finite. Take any $\phi \le f$ and $\alpha \in (0, 1)$, then $f = \limv{n}f_n \ge \phi$. Fix $x \in X$, then if $0 < f(x) < \infty$, $ \exists N \in \nat: f_n(x) \ge f(x) - \varepsilon \ge \alpha \phi (x) \quad \forall n \ge N $ for suitable $\varepsilon$. If $f(x) = 0$, then $0 = f_n(x) \ge f(x) \forall n \in \nat$. If $f(x) = \infty$, then $ \exists N \in \nat: f_n(x) \ge \alpha\phi(x) \quad \forall n \in \nat $ $f_n(x)$ can be arbitrarily large. This allows $ E_n = \bracs{x \in X: f_n(x) \ge \alpha \phi(x)} \quad \bigcup_{n \in \nat}E_n $ Either way, we apply the same arguments to get that $ \limv{n}\int f_n \ge \alpha \int \phi \quad \forall \alpha \in (0, 1) $ and $\limv{n}\int f_n \ge \int \phi$. Since this holds for all $\phi \le f$, we have $ \limv{n}\int f_n \ge \int \limv{n}f_n $