If \(\phi \in L_+^0(X)\) is a simple function of the form \(\phi=\sum_{j=1}^n a_j\mathbf{1}_{{E_j}}\) we define the integral of \(\phi\) with respect to \(\mu\) by \[\color{blue}\int_X \phi(x) d\mu(x)=\sum_{j=1}^na_j\mu(E_j) \in [0, \infty],\] with the convention that \(0 \cdot \infty=0\).
We shall also write \[\int_{X}\phi(x) d\mu(x)=\int \phi(x)d\mu(x)=\int \phi d\mu =\int \phi.\]
Proposition. Let \(\phi\), \(\psi\) be simple functions in \(L^0_{+}(X)\).
If \(c \geq 0\), then \(\int_X c\phi d\mu =c\int_X \phi d\mu\).
\(\int_X (\phi+\psi) d\mu=\int_X \phi d\mu+\int_X \psi d\mu\).
If \(\phi \leq \psi\), then \(\int_X \phi d\mu\leq \int_X \psi d\mu\).
The map \(A \mapsto \int_A \phi d\mu\) is measurable on \(\mathcal M\).
Proof of (a). Note that \(\int_X c\phi d\mu = \sum_{j=1}^n ca_j\mu({E_j})=c\int_X \phi d\mu\). $$\tag*{$\blacksquare$}$$
Proof of (b). We consider \[E_j=\bigcup_{k=1}^{m}E_j \cap F_k, \qquad F_k=\bigcup_{j=1}^{n}E_j \cap F_k.\] since \(X=\bigcup_{j=1}^n E_j=\bigcup_{k=1}^m F_k\) (disjoint unions). Then \[\begin{aligned} \qquad &\int_X \phi d\mu +\int_X \psi d\mu =\sum_{j=1}^na_j \mu(E_j)+\sum_{k=1}^{m}b_k\mu(F_k)\\ &=\sum_{j=1}^{n}\sum_{k=1}^{m}(a_j+b_k)\mu(E_j \cap F_k)=\int_X (\phi+\psi)d\mu. \qquad {\blacksquare} \end{aligned}\]
Proof of (c). If \(\phi \leq \psi\) then \(a_j \leq b_k\) on \(E_j \cap F_k \neq \varnothing\). Then \[\begin{aligned} \int_X\phi d\mu=\sum_{j=1}^{n}\sum_{k=1}^{m}a_j\mu(E_j \cap F_k) \leq \sum_{j=1}^{n}\sum_{k=1}^{m}b_k\mu(E_j \cap F_k)=\int_X \psi d\mu. \quad {\blacksquare} \end{aligned}\]
Proof of (d). For \(A\in \mathcal M\) define \[\nu(A)=\int_A \phi d\mu=\int_X \phi \mathbf{1}_{{A}}d\mu.\] Note that \(\nu\) is countably additive. Indeed, if \((A_k)_{k \in \mathbb{N}}\) is a disjoint sequence in \(\mathcal M\) and \(A=\bigcup_{k=1}^{\infty}A_k\), then \[\nu(A)=\sum_{j=1}^na_j\mu(A \cap E_j)= \sum_{k \in \mathbb{N}}\sum_{j=1}^na_j\mu(A_k \cap E_j) =\sum_{k \in \mathbb{N}}\int_{A_k}\phi d\mu =\sum_{k \in \mathbb{N}}\nu(A_k),\] since \(\mu\) is countably additive. $$\tag*{$\blacksquare$}$$
The integral can be extended to all functions \(f \in L^0_+(X)\) by defining \[{\color{blue}\int_X f(x) d\mu(x)=\sup\Big\{\int_X \phi(x)d\mu(x): 0 \leq \phi \leq f, \text{ and } \phi \text{ is simple}\Big\}.}\]
Remark.
By the previous proposition (see item (c)), the two definitions of \(\int f\) agree when \(f\) is simple, as the family of simple functions over which the supremum is taken includes \(f\) itself.
If \(f \leq g\), then \(\int_X f d\mu \leq \int_X g d\mu\).
For any \(c \geq 0\) we have \(\int_X cf d\mu =c\int_X fd\mu\).
Let \((X, \mathcal M, \mu)\) be a measure space. If \((f_n)_{n \in \mathbb{N}}\) is a sequence in \(L^0_+(X)\) such that \(f_n \leq f_{n+1}\) for all \(n \in \mathbb{N}\) and \[f =\lim_{n \to \infty}f_n=\sup_{n \in \mathbb{N}} f_n,\] then \[\int_X f(x)d\mu(x)=\lim_{n \to \infty}\int_X f_n(x)d\mu(x).\] In other words, \[{\color{blue}\int \lim_{n \to \infty}f_n(x)d\mu(x)=\lim_{n \to \infty}\int_X f_n(x)d\mu(x).}\]
Proof. Since \((\int_X f_n d\mu )_{n \in \mathbb{N}}\) is an increasing sequence of numbers so the limit exists (possibly equal to \(\infty\)). Since \(f_n \leq f\), then \(\int_X f_n d\mu \leq \int_X f d\mu\) for all \(n \in \mathbb{N}\), and consequently \[\lim_{n \to \infty}\int_X f_n d\mu \leq \int_X f d\mu.\] We now show that \[\lim_{n \to \infty}\int_X f_n d\mu \geq \int_X f d\mu.\] Fix \(\alpha \in (0,1)\), and let \(\phi\) be a simple function with \(0 \leq \phi \leq f\). Consider \[E_n=\{x \in X: f_n(x) \geq \alpha \phi(x)\}.\] Then \(E_n \subseteq E_{n+1}\) for any \(n \in \mathbb{N}\) and \(X=\bigcup_{n \in \mathbb{N}}E_n\), hence \[\int_X f_n d\mu \geq \int_{E_n}f_n d\mu \geq \alpha \int_{E_n}\phi d\mu.\]
By the previous proposition (see item (d)) and the continuity of measure \[\nu(A)=\int_A \phi d\mu\] we obtain \[\lim_{n \to \infty}\int_X f_n d\mu \geq \alpha \lim_{n \to \infty}\int_{E_n}\phi d\mu=\alpha \lim_{n \to \infty}\nu(E_n)=\alpha\int_X \phi d\mu.\] Since \(\alpha \in (0,1)\) was arbitrary \[\lim_{n \to \infty}\int_X f_n d\mu \geq \int_X \phi d\mu.\] Since \(\phi\) was arbitrary simple function such that \(0 \leq \phi \leq f\) thus \[\qquad \lim_{n \to \infty}\int_X f_n d\mu \geq \sup_{0 \leq \phi \leq f}\int_X \phi d\mu=\int_X f d\mu. \qquad \tag*{$\blacksquare$}\]
Let \((X, \mathcal M, \mu)\) be a measure space. If \((f_n)_{n \in \mathbb{N}}\) is a finite or infinite sequence in \(L^0_+(X)\) such that \[f=\sum_{n \in \mathbb{N}}f_n,\] then \[\int_X f d\mu =\int_X \sum_{n \in \mathbb{N}}f_n d\mu =\sum_{n \in \mathbb{N}}\int_X f_n d\mu.\]
Proof. First we show that \[\int_X (f_1+f_2) d\mu =\int_X f_1 d\mu +\int_X f_2d\mu .\] We find sequences \((\phi_n)_{n \in \mathbb{N}}\) and \((\psi_n)_{n \in \mathbb{N}}\) of simple functions in \(L^0_+(X)\) that increase to \(f_1\) and \(f_2\), then \((\phi_n+\psi_n)_{n \in \mathbb{N}} \in L^0_+(X)\) and increases to \(f_1+f_2\).
By the (MCT) we conclude \[\begin{gathered} \int_X(f_1+f_2)d\mu \underbrace{=}_{\text{\bf (MCT)}} \lim_{n \to \infty}\int_X (\phi_n+\psi_n)d\mu\\ =\lim_{n \to \infty}\int_X \phi_n d\mu +\lim_{n \to \infty}\int_X \psi_n d\mu \underbrace{=}_{\text{\bf (MCT)}}\int_X f_1 d\mu +\int_X f_2d\mu . \end{gathered}\] Now by induction \[\int_X \sum_{n=1}^Nf_n d\mu =\sum_{n=1}^N\int_X f_n d\mu \quad \text{ for any } \quad N \in \mathbb{N}.\] Letting \(N \to \infty\) and applying (MCT) again we have \[\int_X \sum_{n=1}^{\infty}f_n d\mu=\int_X \lim_{N \to \infty}\sum_{n=1}^Nf_n d\mu =\lim_{N \to \infty}\sum_{n=1}^N\int_X f_n d\mu =\sum_{n=1}^{\infty}\int_X f_n d\mu. \quad \tag*{$\blacksquare$}\]
Proposition. Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^0_+(X)\), then \[\int_X f d\mu =0 \quad \iff\quad f=0 \text{ a.e..}\]
Proof. This is clear if \(f\) is a simple function. If \(f=\sum_{j=1}^n a_j\mathbf{1}_{{E_j}}\) with \(a_j \geq 0\) then \[0=\int_X fd\mu =\sum_{j=1}^n a_j\mu(E_j),\] iff \(a_j=0\) or \(\mu(E_j)=0\) for each \(1\le j\le n\). In general, if \(f \in L^0_+(X)\) and \(f=0\) a.e. and \(\phi\) is a simple function with \(0 \leq \phi \leq f\), then \(\phi=0\) a.e., hence \[\int_X fd\mu =\sup_{0 \leq \phi \leq f}\int_X \phi d\mu =0.\]
Let \((X, \mathcal M, \mu)\) be a measure space. If \((f_n)_{n \in \mathbb{N}}\subseteq L^0_+(X)\) and \(f_n(x)\) increases to \(f(x)\) for a.e. \(x \in X\), then \[\int_X fd\mu =\lim_{n \to \infty}\int_X f_nd\mu .\]
Proof. If \(f_n(x)\) increases to \(f(x)\) for \(x \in E\), where \(\mu(E^c)=0\), then \(f-f\mathbf{1}_{{E}}=0\) a.e. and \(f_n-f_n\mathbf{1}_{{E}}=0\) a.e. By the (MCT) we have \[\int_X fd\mu =\int_X f \mathbf{1}_{{E}}d\mu =\lim_{n \to \infty}\int_X f_n\mathbf{1}_{{E}}d\mu =\lim_{n \to \infty}\int_X f_nd\mu. \qquad \tag*{$\blacksquare$}\]
Let \((X, \mathcal M, \mu)\) be a measure space. If \((f_n)_{n \in \mathbb{N}}\subseteq L^0_+(X)\), then \[\int_X \liminf_{n \to \infty}f_nd\mu \leq \liminf_{n \to \infty}\int_X f_nd\mu.\]
Proof. For each \(k \geq 0\) we have \[\inf_{n \geq k}f_n \leq f_j \quad \text{ for all } \quad j \geq k.\] thus \[\int_X \inf_{n \geq k} f_nd\mu \leq \int_X f_jd\mu \quad \text{ for } \quad j \geq k,\] hence \[\int_X \inf_{n \geq k}f_nd\mu \leq \inf_{j \geq k}\int_X f_jd\mu.\] Letting \(k \to \infty\) by the (MCT) we obtain the claim. $$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a measure space. If \((f_n)_{n \in \mathbb{N}}\subseteq L^0_+(X)\) and \(\lim_{n\to \infty}f_n = f\) a.e., then \[\int_X fd\mu \leq \liminf_{n \to \infty}\int_X f_nd\mu .\]
Proposition. Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^0_+(X)\) and \(\int_X fd\mu <\infty\), then \[\mu(\{x \in X: f(x)=\infty\})=0\] and \[\{x \in X: f(x)>0\}\] is \(\sigma\)-finite.
If \(\int_x f^+d\mu\) and \(\int_X f^{-}d\mu\) are both finite, we say that \(f\) is integrable.
Remark. Since \(|f|=f^++f^-\), then \(f\) is integrable iff \[{\color{purple}\int_X |f(x)|d\mu(x)<\infty}.\]
Proposition. Let \((X, \mathcal M, \mu)\) be a measure space. The set of integrable real-valued functions on \(X\) is a real vector space and the integral is a linear functional on it.
Proof. The first conclusion follows since \(|af+bg| \leq |a||f|+|b||g|\).
It is easy to check that \(\int af =a\int f\) for any \(a \in \mathbb{R}\).
Suppose \(f,g\) are integrable, then \(\int (f+g) =\int f +\int g.\) Indeed, let \(h=f+g\), then \(h^+-h^-=f^+-f^-+g^+-g^-\) and \[h^++f^-+g^-=h^-+f^++g^+,\] \[\int h^++\int f^-+\int g^-=\int h^-+\int g^++\int f^+,\] \[\int h=\int h^+-\int h^-=\int f^+-\int f^-+\int g^+ -\int g^-=\int f+\int g.\tag*{$\blacksquare$}\]
Let \((X, \mathcal M, \mu)\) be a measure space. If \(f:X \to \mathbb{C}\) is measurable function we say that \(f\) is integrable if \[\int_X |f(x)|d\mu(x)<\infty.\] If \(E \in \mathcal M\), then \(f\) is integrable on \(E\) if \(\int_E |f(x)|d\mu(x)<\infty.\)
Remark. Since \(|f| \leq |{\rm Re\; }(f)|+|{\rm Im\; }(f)| \leq 2|f|.\) Thus \(f\) is integrable iff \({\rm Re}(f)\) and \({\rm Im}(f)\) are both integrable. In this case we define \[\int_X fd\mu =\int_X {\rm Re}(f)d\mu +i\int_X {\rm Im}(f)d\mu.\]
Let \((X, \mathcal M, \mu)\) be a measure space. The space of all complex-valued integrable functions will be denoted by \[L^1(X,\mu)=\Big\{f:X\to \mathbb C: f \text{ is measurable and } \int_X |f(x)|d\mu(x)<\infty\Big\}.\] We shall abbreviate \(L^1(X,\mu)\) to \(L^1(X)\) or \(L^1(\mu)\) or \(L^1\).
Proposition. Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^1(X)\), then \[\bigg|\int_X fd\mu \bigg| \leq \int_X |f|d\mu .\]
Proof. This is obvious if \(\int_X fd\mu =0\). Assume that \(f:X \to \mathbb{R}\), and note \[\bigg|\int_X fd\mu \bigg|=\bigg|\int_X f^{+}d\mu -\int_X f^{-}d\mu \bigg| \leq \int_X f^+d\mu+\int_X f^-d\mu=\int_X |f|d\mu.\] If \(f:X \to \mathbb{C}\) and \(\int_X fd\mu \neq 0\), let \[{\color{purple}\alpha={\rm sgn}\left(\int_X fd\mu \right) =\frac{\overline{\int_X fd\mu}}{|\int_X fd\mu|}}.\] Then \[\bigg|\int_X fd\mu \bigg|=\alpha\int_X fd\mu =\int_X \alpha fd\mu\in \mathbb R,\] thus \(\big|\int_X fd\mu \big|={\rm Re}\left(\int_X \alpha f d\mu\right)\) and \[\bigg|\int_X fd\mu \bigg|=\int_X {\rm Re}(\alpha f)d\mu \leq \int_X |{\rm Re}(\alpha f)|d\mu \leq \int_X |\alpha f|d\mu=\int_X |f|d\mu.\tag*{$\blacksquare$}\]
Proposition (Chebyshev’s inequality). Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^1(X)\) and \(\lambda>0\), then \[\mu(E) \leq \frac{1}{\lambda}\int_{E}|f(x)| d\mu(x),\] where \(E=\{x \in X: |f(x)|>\lambda\}.\)
Proposition. Let \((X, \mathcal M, \mu)\) be a measure space.
If \(f \in L^1(X)\), then \(\{x \in X: f(x) \neq 0\}\) is \(\sigma\)-finite.
If \(f,g \in L^1(X)\), then \(|f-g|=0\) a.e. iff \(f-g=0\) a.e. iff \[\int_E f d\mu =\int_E g d\mu \quad \text{ for all } \quad E \in \mathcal M.\]
Proof of (a). Note that \[\begin{aligned} \{x \in X: f(x) \neq 0\} =\{x \in X: |f(x)|>0\} =\bigcup_{n \in \mathbb{N}}A_n, \end{aligned}\] where \[A_n=\{x \in X: |f(x)| \geq {n}^{-1}\}.\] Then we have \(A_n \subseteq A_{n+1}\) and \[\mu\left(\{x \in X: |f(x)|>0\}\right)=\lim_{n \to \infty}\mu(A_n).\] By Chebyshev’s inequality we have \[\mu(A_n) \leq n\int_{A_n}|f(x)|d\mu(x) \leq n\int_X|f(x)|\mu(x)<\infty.\]
Proof of (b). It is clear that \(\int_X |f-g|d\mu =0\) iff \(f=g\) a.e..
If \(\int_X |f-g|d\mu =0\), then \(\int_E f d\mu =\int_E g d\mu\) for any \(E \in \mathcal M\), since \[\bigg|\int_E fd\mu -\int_E g d\mu\bigg| \leq \int_E |f-g|d\mu \leq \int_X |f-g|d\mu =0.\]
If \(\int_E fd\mu =\int_E g d\mu\) for any \(E \in \mathcal M\). If \(u={\rm Re}(f-g)\), \(v={\rm Im}(f-g)\) and if it is false that \(f-g\) a.e., then at least one of \(u^+\), \(u^-\), \(v^+\), \(v^-\) must be nonzero at the set of positive measure. Suppose that \[E=\{x \in X: u^+(x)>0\}\in\mathcal M\] has positive measure. Then \[{\rm Re}\left(\int_E f d\mu -\int_E gd\mu\right)=\int_E u^+d\mu>0,\] since \(u^-=0\) on \(E\). Thus we must have \(f=g\) a.e.. $$\tag*{$\blacksquare$}$$
Remark. This proposition shows that for the purposes of integration it makes no difference if we alter functions on null sets.
One can integrate functions \(f\) that are only defined on a measurable set \(E\) whose complement is null simply by defining \(f\) to be zero (or anything else) on \(E^c\).
In this fashion we can treat \(\overline{\mathbb R}\)-valued functions that are finite a.e. as real-valued functions for the purposes of integration.
With this in mind, we shall find it more convenient to redefine \(L^1(X, \mu)\) to be the set of equivalence classes of a.e.-defined integrable functions on \(X\), where \(f\) and \(g\) are considered equivalent iff \(f = g\) a.e.
This new \(L^1(X, \mu)\) is still a complex vector space (under pointwise a.e. addition and scalar multiplication).
Remark.
Although we shall henceforth view \(L^1(X, \mu)\) as a space of equivalence classes, we shall still employ the notation “\(f\in L^1(X, \mu)\)” to mean that \(f\) is an a.e.-defined integrable function. This minor abuse of notation is commonly accepted and rarely causes any confusion.
There are two advantages of this definition
If \(\overline{\mu}\) is a completion of \(\mu\) there is a natural one-to-one correspondence between \(L^1(X, \mu)\) and \(L^1(X, \overline{\mu})\). We shall identify these spaces.
\(L^1(X, \mu)\) is a metric space with distance function \[\rho(f,g)=\int_X |f-g|d\mu,\] \[\rho(f,g)=0 \quad \iff \quad f=g \text{ a.e.}\]
We shall refer to convergence with respect to this metric as convergence in \(L^1\), thus \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in \(L^1\) iff \(\lim_{n\to \infty}\int_X|f_n-f|d\mu=0\).
Let \((X, \mathcal M, \mu)\) be a measure space. Let \((f_n)_{n \in \mathbb{N}}\subseteq L^1(X)\) be such that
\(f_n \ _{\overrightarrow{n\to \infty}} \ f\) a.e.,
there exists \(g\in L^1(X)\) such that \(g\ge0\) and \(|f_n| \leq g\) a.e. for all \(n \in \mathbb{N}\).
Then \(f \in L^1(X)\) and \[{\color{purple}\int_X fd\mu= \int_X \lim_{n \to \infty}f_nd\mu=\lim_{n \to \infty}\int_X f_nd\mu.}\]
Proof. We can assume (perhaps after redefinition on a null set) that the function \(f\) is measurable. Note that \[|f|=\lim_{n \to \infty}|f_n| \leq g \quad \text{ a.e..}\] Thus \(f \in L^1(X)\), since \[\int_X |f|d\mu \leq \int_X g d\mu<\infty.\] By taking real and imaginary parts it suffices to assume that \(f_n\) and \(f\) are real-valued function. Then \[|f_n| \leq g \quad \iff \quad -g \leq f_n \leq g \quad \text{ a.e.,}\] which yields \[g+f_n \geq 0 \quad \text{a.e.}\quad \text{ and } \quad g-f_n \geq 0 \quad \text{a.e.}\]
Thus by Fatou’s Lemma we obtain \[\int_X gd\mu+\int_X f d\mu \leq \liminf_{n \to \infty}\int_X (g+f_n)d\mu =\int_X g d\mu+\liminf_{n \to \infty}\int_X f_nd\mu,\] and \[\int_X gd\mu-\int_X f d\mu \leq \liminf_{n \to \infty}\int_X (g-f_n)d\mu =\int_X g d\mu-\limsup_{n \to \infty}\int_X f_nd\mu.\] Thus \[\int_X f d\mu \leq \liminf_{n \to \infty}\int_X f_nd\mu\le \limsup_{n \to \infty}\int_X f_nd\mu \le \int_X f d\mu,\] and consequently \[\qquad \int_X f d\mu =\lim_{n \to \infty}\int_X f_nd \mu. \qquad \tag*{$\blacksquare$}\]
Let \((X, \mathcal M, \mu)\) be a measure space. Let \((f_n)_{n \in \mathbb{N}}\subseteq L^1(X)\) be such that \[\sum_{n \in \mathbb{N}}\int_X |f_n|d\mu <\infty.\] Then the series \[\sum_{n \in \mathbb{N}}f_n\] converges a.e. to a function in \(L^1(X)\) and \[\int_X \sum_{n \in \mathbb{N}}f_nd\mu = \sum_{n \in \mathbb{N}}\int_X f_n d\mu.\]
Proof. By the (MCT) we have \[\int_X \sum_{n \in \mathbb{N}}|f_n|d\mu = \sum_{n \in \mathbb{N}}\int_X |f_n|d\mu <\infty.\] So the function \(g =\sum_{j \in \mathbb{N}}|f_j| \in L^1(X)\) thus \[g(x) =\sum_{j \in \mathbb{N}}|f_j(x)|<\infty \quad \text{a.e.}\] Thus the series \(\sum_{n \in \mathbb{N}}f_n(x)\) converges a.e. For each \(N\in\mathbb N\) we have \[\Big|\sum_{j=1}^Nf_j(x)\Big| \leq g(x) \quad \text{ a.e.}\] We apply (DCT) and we obtain \[\qquad\int_X \sum_{n \in \mathbb{N}}f_nd\mu =\int_X \lim_{N \to \infty}\sum_{n=1}^N f_nd\mu =\lim_{N \to \infty}\int_X \sum_{n=1}^{N}f_nd\mu. \qquad \tag*{$\blacksquare$}\]