> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], and $f: X \to \real$ be a $\cm$-[[Measurable Function|measurable function]]. Then the **positive** and **negative** parts of $f$ are > $ > f^+ = \max(f, 0) \quad f^- = -\min(f, 0) \quad f = f^+ - f^- > $ > where $f^+, f^- \in L^+$. If the [[Integral|integrals]] $\int f^+, \int f^- < \infty$ are finite (or if $\int|f| < \infty$), then $f$ is **integrable**, and > $ > \int f d\mu = \int f^+ d\mu - \int f^-d\mu > $ > [!definition] > > Let $f: X \to \real$ be a $\cm$-measurable function. Then $f$ is **extended** $\mu$-integrable if $\int f^+ d\mu < \infty$ or $\int f^- < \infty$, and > $ > \int f = \int f^+ d\mu - \int f^- d\mu > $ > [!definition] > > Let $f: X \to \complex$ be a $\cm$-measurable function. Then $f$ is **integrable** if $\int |f| < \infty$ and > $ > \int f d\mu = \int\re{f}d\mu + i\int\im{f}d\mu > $ > Since $|f| \le |\re f| + |\im f| \le 2|f|$, $f \in L^1(\mu)$ if and only if $\re f \in L^1(\mu), \im f \in L^1(\mu)$. > [!definition] > > Let $(X, \cm)$ be a measurable space and $\nu$ be a [[Signed Measure|signed measure]] on $(X, \cm)$. Let $\nu = \nu^+ - \nu^-$ be the [[Jordan Decomposition Theorem|Jordan decomposition]] of $\nu$, then define > $ > L^1(\nu) = L^1(\nu^+) \cap L^1(\nu^-) > $ > and for any $\cm$-measurable function $f \in L^1(\nu)$, > $ > \int f d\nu = \int f d\nu^+ - \int f d\nu^- > $ > [!definition] > > Let $(X, \cm, \mu)$ be a measure space and $f: X \setminus N \to \real|\complex$ be a [[Measurable Function|measurable function]] defined [[Almost Everywhere|almost everywhere]]. Then extending $f$ to $X$ by $f(x) = 0 \forall x \in N$ yields a $\cm$-measurable function. > > Let $f \sim g$ be [[Equivalence Relation|equivalent]] if $f = g$ a.e., and denote $[f]$ as its [[Equivalence Class|equivalence class]]. $L^1(\mu)$ is the space of the equivalence classes of all integrable functions. > > When mentioning $f \in L^1(\mu)$, it refers to $f: [f] \in L^1(\mu)$ where $f$ is an a.e.-defined integrable function. # As a Vector Space > [!theorem] > > Let $f, g \in L^1(\mu)$ and $\alpha \in \real$, then > $ > \int (f + g) = \int f + \int g \quad \int \alpha f = \alpha \int f > $ > *Proof*. First suppose that $f, g$ are real-valued. Let $h = f + g$, then $h^+ - h^- = f^+ - f^- + g^+ - g^-$ and > $ > \begin{align*} > h^+ + f^- + g^- &= h^- + f^+ + g^+ \\ > \int h^+ + \int f^{-} + \int g^{-} &= \int h^{-} + \int f^{+} + \int g^{+} \\ > \int h^+ - \int h^-&= \int f^+ - \int f^{-} + \int g^+ - \int g^- \\ > \int h&= \int f + \int g > \end{align*} > $ > If $\alpha \ge 0$, then $\alpha f = \alpha f^+ - \alpha f^-$ and $\int f = \alpha \int f$. If $\alpha < 0$, then > $ > \alpha f = (\alpha f)^{+} - (\alpha f)^{-} = > (-\alpha)f^{-} - (-\alpha)f^{+} > $ > and $\int f = \int \alpha f$. > > Now suppose that $f, g$ are complex-valued, then > $ > \begin{align*} > \int f + g &= \int \re{f + g} + i\int \im{f + g} \\ > &= \int \re{f} + \int\re{g} + i\int\im{f} + i\int\im{g} \\ > &= \int f + \int g > \end{align*} > $ > Let $\alpha = a + bi \in \complex$, then > $ > \begin{align*} > \int \alpha f &= \int \re{\alpha f} + i\int\im{\alpha f} \\ > &= \int \paren{a\re{f} - b\im{f}} + i\int\paren{a\im{f} + b\re{f}} \\ > &= a\paren{\int\re{f} + i\int \im{f}} + b\paren{-\int\im{f} +i\int\re{f}}\\ > &= a\paren{\int\re{f} + i\int \im{f}} + bi\paren{\int\re{f}+\int\im{f}} \\ > &= \alpha\int f > \end{align*} > $ > [!theorem] > > The space $L^1(\mu)$ of integrable functions is a [[Real Numbers|real]]-valued [[Vector Space|vector space]]. The [[Integral|integral]] is a [[Linear Functional|linear functional]] on $L^1(\mu)$. > > *Proof.* Let $f, g \in L^1(\mu)$, then > $ > \int |f + g| \le \int |f| + \int |g| < \infty > $ > and $f + g \in L^1(\mu)$. For any scalar $\alpha \in \real$, > $ > \int |\alpha f| = |\alpha|\int |f| < \infty > $ > and $\alpha f \in L^1(\mu)$. Since $L^1(\mu)$ is closed under addition and scalar multiplication, $L^1(\mu)$ forms a vector space. > > For any $f, g \in L^1(\mu)$, $\alpha, \beta \in \real$, > $ > \begin{align*} > \int (\alpha f + \beta g) = \int \alpha f + \int \beta g =\alpha \int f + \beta \int g > \end{align*} > $ > and the integral is a linear functional on $L^1(\mu)$. # As a Metric Space > [!theorem] > > Let $f \in L^1$, then $\abs{\int f} \le \int |f|$. > > *Proof*. If $f$ is real-valued, then > $ > \abs{\int f} = \abs{\int f^+ - \int f^-} \le \int f^+ + \int f^{-} = \int |f| > $ > If $f$ is complex-valued and $\int f \ne 0$, then take $\alpha = \ol{\sgn(\int f)}$, and we have $\abs{\int f} = \alpha \int f = \int \alpha f$. Since $\int \alpha f$ is real, > $ > \abs{\int f} = \re{\int \alpha{f}} = \int\re{\alpha f} \le \int |\re{\alpha f}| \le \int |\alpha f| = \int |f| > $ > [!theorem] > > Let $f \in L^1$, then $\bracs{x: f(x) \ne 0}$ is $\sigma$-finite. For any $f, g \in L^1$, the following are equivalent: > - $\int_E f = \int_E g$ for any $E \in \cm$. > - $\int |f - g| = 0$. > - $f = g$ [[Almost Everywhere|a.e.]] > > *Proof*. First suppose that $f$ is real-valued, then > $ > \bracs{x: f(x) \ne 0} = \bracs{x: f(x) > 0} \cup \bracs{x: -f(x) > 0} > $ > Since $f \in L^1$, $\int f^+ < \infty$ and $\int f^- < \infty$. We have reduced this problem to a function $g \in L^+$. As, > $ > \bracs{x: g(x) > 0} = \bigcup_{n \in \nat}\bracs{x: g(x) > 1/n} > $ > If $\bracs{x: g(x) > 0}$ is not $\sigma$-finite, then there must exist $n \in \nat$ where $\bracs{x: g(x) > 1/n}$ is not $\sigma$-finite. Otherwise we can partition each $\bracs{x: g(x) > 1/n} = \bigcup_{k \in \nat}E_{n, k}$ where each $\mu(E_{n, k}) < \infty$, and taking $F_{n, k} = E_{n, k} \setminus \bigcup_{j = 1}^{n - 1}\bracs{x: g(x) > 1/n}$ yields a disjoint union $X = \bigcup_{n \in \nat}\bigcup_{k \in \nat}F_{n, k}$ where each $\mu(F_{n, k}) < \infty$, and $X$ is $\sigma$-finite. > > If $\bracs{x: g(x) > 1/n}$ is not $\sigma$-finite, then > $ > \int g \ge \int_{\bracs{x: g(x) > 1/n}}\frac{1}{n} = \infty > $ > and $\int g = \infty$. If $f \in L^1$, then $\int f^+, \int f^- < \infty$ and each $\bracs{x: f(x) > 0}$, $\bracs{x: f(x) < 0}$ is $\sigma$-finite. Therefore $\bracs{x: f(x) \ne 0}$ is $\sigma$-finite. > > Now suppose that $f$ is complex-valued. $f \in L^1$ implies that $\re f, \im f \in L^1$, and $\bracs{x: \re {f(x)} \ne 0}$ and $\bracs{x: \im{f(x)} \ne 0}$ are both $\sigma$-finite. Since > $ > \bracs{x: f(x) \ne 0} = \bracs{x: \re{f(x)} \ne 0} \cup \bracs{x: \im{f(x)} \ne 0} > $ > we have $\bracs{x: f(x) \ne 0}$ being $\sigma$-finite as well. > > Let $f, g \in L^1$, then $\int |f - g| = 0$ if and only if $|f - g| = 0$ a.e., if and only if $f = g$ a.e. > > Suppose that $\int |f - g| = 0$, then for any $E \in \cm$, > $ > \abs{\int_{E}f - \int_{E}g} = \abs{\int_E f - g} \le \int_E|f - g| \le \int|f - g| = 0 > $ > and $\int_E f = \int_E g$. Now suppose that $f = g$ a.e., and let $N \supset \bracs{x: f(x) \ne g(x)}$ be a [[Null Set|null set]], then for any $E \in \cm$, > $ > \begin{align*} > \int_{E} f &= \int_{E \setminus N}f + \int_{E \cap N}f \\ > &= \int_{E \setminus N}g \\ > &= \int_{E \setminus N}g + \int_{E \cap N}g \\ > &= \int_{E} g > \end{align*} > $ > Since $f = g$ a.e. is equivalent to $\int|f - g| = 0$, all statements are equivalent. > [!theorem] > > $L^1(\mu)$ is a [[Metric Space|metric space]] with distance function $\rho([f], [g]) = \int |f - g|$. > > *Proof*. Let $[f], [g] \in L^1(\mu)$. We first verify that the function is well-defined. Let $f_1, f_2 \in [f]$ and $g_1, g_2 \in [g]$, then > $ > \begin{align*} > \int |f_1 - g_1| &= \int |f_1 - f_2 + f_2 - g_2 + g_2 - g_1| \\ > &\le \int |f_1 - f_2| + \int |f_2 - g_2| + \int |g_2 - g_1| \\ > &= \int |f_2 - g_2| > \end{align*} > $ > Since the argument is symmetric, we have $\int|f_1 - g_2| = \int|f_2 - g_2|$. > > Now, $\int |f - g| = \int |g - f|$, and for any $[h] \in L^1(\mu)$, > $ > \begin{align*} > \int |f - h| &= \int|f - g + g - h| \\ > &\le \int |f - g| + \int |g - h| > \end{align*} > $ > and the [[Triangle Inequality|triangle inequality]] holds. > > Finally, $\int |f - g| = 0$ if and only if $f = g$ a.e., which only happens when $[f] = [g]$. > [!theorem] > > Let $\seq{f}$ be a [[Sequence|sequence]] in $L^1$ such that $\sum_{n \in \nat}\int |f_n| < \infty$. Then $\sum_{n \in \nat}f_n$ converges [[Almost Everywhere|a.e.]] to a function in $L^1$, and $\int \sum_{n \in \nat}f_n = \sum_{n \in \nat}\int f_n$. > > *Proof.* Since each $|f_n| \in L^+$, $\int \sum_{n \in \nat}|f_n| = \sum_{n \in \nat}\int|f_n|$ by the [[Monotone Convergence Theorem]]. Therefore $\sum_{n \in \nat}f_n$ is dominated by $\sum_{n \in \nat}|f_n|$. Since $\int\sum_{n \in \nat}|f_n| = \sum_{n \in \nat}\int|f_n| < \infty$, $\sum_{n \in \nat}|f_n|$ is finite and convergent a.e., so $\sum_{k = 1}^n f_n \to \sum_{n \in \nat}f_n$ a.e. Applying the [[Dominated Convergence Theorem]], we have $\int \sum_{n \in \nat}f_n = \sum_{n \in \nat}\int f_n$. > [!theorem] > > Let $f \in L^1(\mu)$ and $\varepsilon > 0$, then there exists a [[Simple Function|simple function]] $\phi = \sum_{j = 1}^{n}a_j\chi_{E_j}$ such that $\int |f - \phi|d\mu < \varepsilon$. In other words, integrable simple functions are [[Dense|dense]] in $L^1$. > > If $\mu$ is a [[Lebesgue-Stieltjes Measure]] on $\real$, then the sets $E_j$ can be taken to be finite unions of open intervals. Moreover, there is a [[Continuity|continuous]] function $g$ vanishing outside of a bounded interval such that $\int|f - g|d\mu < \varepsilon$. > > *Proof*. Since $f \in L^1(\mu)$ is a [[Measurable Function|measurable function]], there exists a sequence $\seq{\phi_n}$ such that $|f| \ge |\phi_n|$ for all $n$, and $\phi_n \to f$ [[Pointwise Convergence|pointwise]]. As $\seq{\phi_n}$ is dominated by $|f|$, $\int f = \limv{n}\int \phi_n$ by the [[Dominated Convergence Theorem]]. Therefore there exists $\phi_n$ such that $\int |f - \phi|d\mu < \varepsilon$. > > Let $\phi = \sum_{j = 1}^{n}a_j\chi_{E_j}$ such that $\int|f - \phi|d\mu < \varepsilon/2$. Remove the term $0\chi_{E_j}$, then $\phi \in L^1$, we have $\mu(E_j) < \infty$ for each $E_j$. If $\mu$ is a Lebesgue-Stieltjes measure, then for each $E_j$ and $\varepsilon > 0$, > $ > \exists U_j \text{ a finite union of open intervals}: \mu(E_j \Delta U_j) < \frac{\varepsilon}{|a_j|2^{j+1}} > $ > and > $ > \begin{align*} > \abs{\chi_{E_j} - \chi_{U_j}} &= \abs{\chi_{E_j \setminus U_j} + \chi_{E_j \cap U_j} - \chi_{E_j \cap U_j} - \chi_{U_j \setminus E_j}} \\ > &= |\chi_{E_j \setminus U} - \chi_{U_j \setminus E_j}| \\ > &= \begin{cases} > |1| & x \in E_j \setminus U_j \\ > |-1| &x \in U_j \setminus E_j \\ > 0 & x \in X \setminus (E_j \Delta U_j) > \end{cases} \\ > &= \chi_{E_j \Delta U_j} > \end{align*} > $ > Let $\psi = \sum_{j = 1}^{n}a_j\chi_{U_j}$, then > $ > \begin{align*} > \int \abs{\phi - \psi} &= \int\abs{\sum_{j = 1}^{n}a_j(\chi_{E_j} - \chi_{U_j})} \\ > &\le \int \sum_{j = 1}^{n}\abs{a_j(\chi_{E_j} - \chi_{U_j})} \\ > &= \sum_{j = 1}^{n}\int|a_j\chi_{E_j \Delta U_j}| \\ > &= \sum_{j = 1}^{n}|a_j|\mu(E_j \Delta U_j) \\ > &< \sum_{j = 1}^{n}\frac{\varepsilon}{2^{j+1}} \\ > &< \varepsilon/2 > \end{align*} > $ > and > $ > \int |f - \psi| \le \int |f - \phi| + \int |\phi - \psi| < \varepsilon > $ > While $\psi = \sum_{j = 1}^{n}a_j\chi_{U_j}$ is not in standard form, it is still a sum of simple functions, and therefore simple. > > Fix $\varepsilon > 0$ and let $I = (a, b)$ be a bounded interval. Let $F$ be the function associated with $\mu$, then $F$ is increasing and right-continuous and > $ > \begin{align*} > \exists \delta > 0: F(a + \delta) - F(a) &< \varepsilon/2 \\ > F(b) - F(b - \delta) &< \varepsilon/2 > \end{align*} > $ > Then > $ > \begin{align*} > \mu((a, b)) &= \mu((a, a + \delta]) + \mu((a + \delta, b - \delta]) + \mu((b - \delta, b)) > \end{align*} > $ > and > $ > \mu((a, a + \delta]) < \varepsilon/2 \quad \mu((b - \delta, b)) \le \mu((b - \delta, b]) < \varepsilon/2 > $ > which allows the construction > $ > g(x) = \begin{cases} > 1 &x \in [a + \delta, b - \delta] \\ > \frac{x - a}{\delta}&x \in (a, a + \delta) \\ > -\frac{x - (b - \delta) - \delta}{\delta} &x \in (b - \delta, b) \\ > 0 &x \in (a, b)^c > \end{cases} > $ > Then $g$ is continuous, vanishes outside of $(a, b)$, and $\chi_{(a + \delta, b - \delta]} \le g \le \chi_{(a, b)}$. So > $ > \begin{align*} > \int |\chi_{(a, b)} - g| &= \int \chi_{(a, b)} - g \\ > &\le \int \chi_{(a, b)} - \chi_{(a + \delta, b - \delta]} \\ > &= \int \chi_{(a, a + \delta] \cup (b - \delta, b)} \\ > &\le \int \chi_{(a, a + \delta)} + \chi_{(b - \delta, b]} \\ > &= \mu((a, a + \delta]) + \mu((b - \delta, b]) \\ > &< \varepsilon > \end{align*} > $ > > Let $U = \bigcup_{i = 1}^{n}I_n$ where each $I_n$ is a bounded interval. Assume that the union is disjoint[^1], then $\chi_{U} = \sum_{i = 1}^{n}\chi_{I_i}$. Let $\varepsilon > 0$ and let $g_i$ be a continuous function vanishing outside of $I_i$ such that $\int |\chi_{I_i} - g_i| < \varepsilon/n$. Take $g = \sum_{i = 1}^{n}g_i$, then $g$ is continuous, vanishes outside of $U$ and > $ > \begin{align*} > \int \abs{\chi_U - g} &= \int\abs{\sum_{i = 1}^{n}\chi_{I_i} - g_i} \\ > &\le \int\sum_{i = 1}^{n}\abs{\chi_{I_i} - g_i} \\ > &= \sum_{i = 1}^{n}\int\abs{\chi_{I_i} - g_i} \\ > &< \sum_{i = 1}^{n} \varepsilon/n = \varepsilon > \end{align*} > $ > Since $U$ is a union of bounded open intervals, $U$ is contained in a bounded open interval, and $g$ vanishes outside of that interval. > > Fix $\varepsilon > 0$ and let $\psi = \sum_{j = 1}^{n}a_j\chi_{U_j}$ such that each $U_j$ is a finite union of open intervals, and $\int |f - \psi| < \varepsilon/2$. Let $g_j$ be a continuous function vanishing outside of $U$ such that $\int |\chi_{U_j} - g_j| < \frac{\varepsilon}{2|a_j|n}$ (or simply $\frac{\varepsilon}{2n}$ if $|a_j| = 0$). Take $g = \sum_{j = 1}^{n}a_jg_j$, then $g$ is continuous, vanishes outside of $\bigcup_{j = 1}^{n}U$, and > $ > \begin{align*} > \int |\psi - g| &= \int \abs{\sum_{j = 1}^{n}a_j\chi_{U_j} - a_jg_j} \\ > &\le \int\sum_{j = 1}^{n}\abs{a_j(\chi_{U_j} - g_j)} \\ > &= \sum_{j = 1}^{n}|a_j|\int|\chi_{U_j} - g_j| \\ > &< \sum_{j = 1}^{n}\frac{\varepsilon}{2n} = \varepsilon/2 > \end{align*} > $ > Which leaves > $ > \int |f - g| \le \int |f - \psi| + \int|\psi - g| < \varepsilon > $ [^1]: See [[Open Interval Shenanigans]].