> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], then $L^+$ is the space of $\cm$-[[Measurable Function|measurable functions]] from $X$ to $[0, \infty]$. > [!definition] > > Let $\phi \in L^+$ be a [[Simple Function|simple function]] with standard representation $\phi = \sum_{i = 1}^{n}a_i\chi_{E_i}$, then the **integral** of $\phi$ with respect to $\mu$ is > $ > \int\phi(x)d\mu(x) = \int \phi d\mu = \sum_{i = 1}^{n}a_j \mu(E_j) > $ > with $0 \cdot \infty = 0$. Let $A \in \cm$, then $\phi \cdot \chi_A$ is also simple, and the *definite* integral is > $ > \int_A \phi(x) d\mu(x) = \int_A \phi d\mu = \int \phi \cdot \chi_A d\mu > $ > [!definition] > > Let $f \in L^+$ be any measurable function, then the [[Upper and Lower Approximations|lower approximation]] of $f$ using simple functions > $ > \mu(f) = \int f d\mu = \sup\bracs{\int \phi d\mu: 0 \le \phi \le f, \phi \text{ simple}} > $ > is the **integral** of $f$. > [!theorem] Strict Reduction > > Let $f \in L^+$ be any measurable function, then we can approximate $f$ strictly with > $ > \ul\Sigma_{x} = \bracs{\phi \text{ simple}: 0 \le \phi \le f, \phi(x) = f(x) \Leftrightarrow f(x) = 0} > $ > then $\int f = \sup_{\phi \in \ul\Sigma_x}\int \phi$. > > *Proof*. Since $\ul\Sigma_x$ is a subset of the collections used to approximate $f$ from below, $\int f \ge \sup_{\phi \in \ul\Sigma_x}\int \phi$. > > Let $0 \le \phi \le f$, then for any $\alpha < 1$, $\phi(x) > 0 \Rightarrow \alpha \phi(x) < \phi(x)$. If $f(x) > 0$, then either $\phi(x) > 0$ or $\phi(x) = 0$. In both cases, $\alpha \phi (x) < f(x)$. If $\phi(x) < f(x)$, then $f(x) > 0$. Therefore $\alpha \phi \in \ul\Sigma_x$. > > We now have > $ > \sup_{\alpha < 1}\int \alpha\phi = \sup_{\alpha < 1}\alpha \int \phi = \int \phi > $ > and therefore $\sup_{\phi \in \ul\Sigma_x}\int \phi \ge \int f$. > [!theorem] > > Let $f \in L^+$, then $\int f = 0 \Leftrightarrow f = 0$ [[Almost Everywhere|a.e.]] > > *Proof*. Let $f \in L^+$ such that $f = 0$ a.e. Let $N \in \cm: \mu(N) = 0$ be a null set where $N \supseteq \bracs{x \in X: f(x) \ne 0}$. Let $\phi \le f$ be a simple function, then > $ > \phi = 0 \cdot \chi_{X \setminus \cn} + \sum_{j = 1}^{n}a_j \chi_{E_j} \quad E_j \subseteq N > $ > since $\phi(x) > 0 \Rightarrow x \in N$. So > $ > \int \phi = 0 + \sum_{j = 1}^{n}a_j\mu(E_j) = 0 > $ > and $\int f = 0$. > > Now suppose that $\int f = 0$, then $\int \phi = 0 \forall \phi \le f$. Let $E_j = \bracs{x \in X: f(x) > 1/j}$, then > $ > \forall a > 0, \exists N \in \nat, 1/N < a: a \in E_N > $ > and $\bigcup_{j \in \nat}E_j = \bracs{x \in X: f(x) > 0}$. Since $\frac{1}{j}\chi_{E_j} \le f$, > $ > \int \frac{1}{j}\chi_{E_j} = \frac{1}{j}\mu(E_j) = 0 \Rightarrow \mu(E_j) = 0 \quad \forall j \in \nat > $ > which gives $\mu\paren{\bigcup_{j \in \nat}E_j} \le \sum_{j \in \nat}\mu(E_j) = 0$ and $f = 0$ a.e. > [!theorem] > > Let $f \in L^+$ such that $\int f < \infty$. Then $\bracs{x: f(x) = \infty}$ is a null set and $\bracs{x: f(x) > 0}$ is $\sigma$-finite. > > *Proof*. Let $E = \bracs{x: f(x) = \infty}$, then $\alpha \chi_{E} \le f \forall \alpha \ge 0$. This means that $\mu(E) \le 1/\alpha \forall \alpha \ge 0$ and $\mu(E) = 0$. > > Let $F = \bracs{x: f(x) > 0}$ and $F_n = \bracs{x: f(x) > 1/n}$. Then $F = \bigcup_{n \in \nat}F_n$. Suppose that $F = \bracs{x: f(x) > 0}$ is not sigma finite, then there must exist $F_k$ where $F_k$ is not sigma finite. If not, then we can simply partition each $F_k$ as a countable union of finite sets, and combine them to create $F$ as a countable union of finite sets. This gives > $ > \int \frac{1}{k}\chi_{F_k} = \frac{\infty}{k} \Rightarrow \int f = \infty > $ > Therefore $F$ cannot be sigma finite. > [!theorem] > > Let $f: X \times [a, b] \to \complex$ ($[a, b]$ is bounded) such that $f(\cdot, t): X \to \complex$ is [[Integrable Function|integrable]], and let $F(t) = \int_X f(x, t)d\mu(x)$. > > If there exists $g \in L^1(\mu)$ such that $|f(x, t)| \le g(x)$ for all $x \in X, t \in [a, b]$, and $\lim_{t \to t_0}f(x, t) = f(x, t_0)$ for all $x$, then the [[Limit|limit]] may be interchanged with the integral: > $ > \lim_{t \to t_0}F(t) = \lim_{t \to t_0}\int_X f(x, t)d\mu(x) = \int_xf(x, t_0)d\mu(x) = F(t_0) > $ > If $f(x, \cdot)$ is [[Continuity|continuous]] for all $x$, then $F$ is also continuous. > > If $\frac{\partial f}{\partial t}$ exists, and there exists $g \in L^1(\mu)$ such that $\abs{\frac{\partial f}{\partial t}(x, t)} \le g(x)$ for all $x, t$, then $F$ is [[Derivative|differentiable]] and the [[Derivative|derivative]] may be interchanged with the integral: > $ > \frac{dF}{dt} = \frac{d}{dt}\int_X f(x, t)d\mu(x) = \int_X \frac{\partial f}{\partial t}(x, t)d\mu(x) > $ > > *Proof*. Let $\seq{t_n} \subset [a, b]$ be any [[Sequence|sequence]] such that $t_n \to t_0$, then since $f(x, t_n)$ is dominated by $g$ for all $t_n$ and $f(x, t_n) \to f(x, t_0)$, by the [[Dominated Convergence Theorem]], > $ > \limv{n}\int_X f(x,t_n)d\mu(x) = \int\limv{n}f(x, t_n)d\mu(x) = \int f(x,t_0)d\mu(x) > $ > Since this holds for any $t_n \to t_0$, we have > $ > \lim_{t \to t_0}\int_X f(x, t)d\mu(x) = \int f(x, t_0)d\mu(x) > $ > If $f(x, \cdot)$ is continuous for all $x$, then for any $t_0 \in [a, b]$, > $ > \lim_{t \to t_0}f(x, t) = f(x, t_0) > $ > which means that > $ > \lim_{t \to t_0}\int_X f(x, t)d\mu(x) = \int f(x, t_0)d\mu(x) \quad \forall t_0 \in [a, b] > $ > and $F$ is continuous. > > Suppose that $\frac{\partial f}{\partial t}$ exists for all $x, t$, then > $ > \frac{\partial f}{\partial t}(x, t_0) = \lim_{t \to t_0}\frac{f(x, t_0) - f(x, t)}{t_0 - t} > $ > meaning that > $ > \frac{\partial f}{\partial t}(x, t_0) = \limv{n}\frac{f(x, t_0) - f(x, t_n)}{t_0 - t_n} > $ > for any sequence $\seq{t_n} \subset [a, b]$ where $t_n \to t_0$ ($t_n \ne t_0$). Since each $f(x, t_0)$ is integrable and therefore [[Measurable Function|measurable]], $\frac{\partial f}{\partial t}(x, t_0)$ is a limit of measurable functions, which is also measurable. As $\frac{\partial f}{\partial t}(x, t_0)$ is dominated by $g(x)$, $\frac{\partial f}{\partial t}(x, t)$ is integrable for each $t \in [a, b]$. > > Denote > $ > h_n(x) = \frac{f(x, t_0) - f(x, t_n)}{t_0 - t_n} > $ > then by the [[Mean Value Theorem|mean value theorem]], for any $x \in X$, there exists $c$ between $t_0$ and $t_n$ such that $h_n(x) = \frac{\partial f}{\partial t}(x, c)$. This means that > $ > \forall x \in X, \exists c \in [a, b]: \abs{h_n(x)} = \abs{\frac{\partial f}{\partial t}(x, c)} \le \sup_{t \in [a, b]}\abs{\frac{\partial f}{\partial t}(x, t)} \le g(x) > $ > and $h_n(x)$ is dominated by $g(x)$ as well. By the [[Dominated Convergence Theorem]], > $ > \begin{align*} > \frac{dF}{dt}(t_0) &= \limv{n}\frac{F(t_0) - F(t_n)}{t_0 - t_n} \\ > &= \limv{n}\frac{\int_X f(x, t_0)d\mu(x) - \int_X f(x, t_n)d\mu(x)}{t_0 - t_n} \\ > &= \limv{n}\int_X\frac{f(x, t_0) - f(x, t_n)}{t_0 - t_n}d\mu(x) \\ > &= \int_X \limv{n}\frac{f(x, t_0) - f(x, t_n)}{t_0 - t_n}d\mu(x) \\ > &= \int_X \frac{\partial f}{\partial t}(x, t_0)d\mu(x) > \end{align*} > $ > Since the above holds for any sequence $t_n \to t_0$, > $ > \begin{align*} > \frac{dF}{dt}(t_0) &= \lim_{t \to t_0}\frac{F(t_0) - F(t)}{t_0 - t} \\ > &= \int_X \lim_{t \to t_0}\frac{f(x, t_0) - f(x, t)}{t_0 - t}d\mu(x) \\ > &= \int_X\frac{\partial f}{\partial t}(x, t_0)d\mu(x) > \end{align*} > $ > The derivative $\frac{dF}{dt}$ therefore exists at every $t_0 \in [a, b]$, and $\frac{dF}{dt} = \int_X \frac{\partial f}{\partial t}(x, t)d\mu(x)$.