Measure Theory/Integration

Let $$(X,\sigma,\mu)$$ be a $$\sigma$$-finite measure space. Suppose $$s$$ is a positive simple measurable function, with $$s=\displaystyle\sum_{i=1}^3y_i\chi_{A_i}$$; $$A_i\in\sigma$$ are disjoint.

Define $$\displaystyle\int_Xs~d\mu=\sum y_i\mu(A_i)$$

Let $$f:X\to\overline{\mathbb{R}}$$ be measurable, and let $$f\geq 0$$.

Define $$\displaystyle\int_Xf~d\mu=\sup\{\int_Xs~d\mu=\sum y_i\mu(A_i)|s\text{ simple },s\geq0,s\leq f\}$$

Now let $$f$$ be any measurable function. We say that $$f$$ is integrable if $$f^+$$ and $$f^-$$ are integrable and if $$\displaystyle\int_Xf^+~d\mu,\int_Xf^-~d\mu<\infty$$. Then, we write

$$\displaystyle\int_Xf~d\mu=\int_Xf^+~d\mu-\int_Xf^-~d\mu$$

The class of measurable functions on $$X$$ is denoted by $$\mathcal{L}^1(X)$$

For $$0<p<\infty$$, we define $$\mathcal{L}^p$$ to be the collection of all measurable functions $$f$$ such that $$|f|^p\in\mathcal{L}^1$$

A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.

Properties
Let $$(X,\sigma,\mu)$$ be a measure space and let $$f,g$$ be measurable on $$X$$. Then
 * 1) If $$f\leq g$$, then $$\displaystyle\int_Xfd\mu\leq\int_Xgd\mu$$
 * 2) If $$A,B\in\sigma$$, $$A\subset B$$, then $$\displaystyle\int_Afd\mu\leq\int_Bfd\mu$$
 * 3) If $$f\geq 0$$ and $$c\geq 0$$ then $$\displaystyle\int_Xcfd\mu=c\int_Xfd\mu$$
 * 4) If $$E\in\sigma$$, $$\mu(E)=0$$, then $$\displaystyle\int_Efd\mu=0$$, even if $$f(E)=\{\infty\}$$
 * 5) If $$E\in\sigma$$, $$f(E)=\{0\}$$, then $$\displaystyle\int_Efd\mu=0$$, even if $$\mu(E)=\infty$$

Proof

Monotone Convergence Theorem
Suppose $$f_n\geq 0$$ and $$f_n$$ are measurable for all $$n$$ such that
 * 1) $$f_1(x)\leq f_2(x)\leq\ldots$$ for every $$x\in X$$
 * 2) $$f_n(x)\to f(x)$$ almost everywhere on $$X$$

Then, $$\displaystyle\int_Xf_nd\mu\to\int_Xfd\mu$$

Proof

$$\displaystyle\int_Xf_nd\mu$$ is an increasing sequence in $$\mathbb{R}$$, and hence, $$\displaystyle\int_Xf_nd\mu\to\alpha\in\overline{\mathbb{R}}$$ (say). We know that $$f$$ is measurable and that $$f\geq f_n\forall n$$. That is,

$$\displaystyle\int_Xf_1d\mu\leq\int_Xf_2d\mu\leq\ldots\int_Xf_nd\mu\leq\ldots\int_Xfd\mu$$

Hence, $$\displaystyle\alpha\leq\int_Xfd\mu=\sup\{\int_Xsd\mu:s\text{ is simple },0\leq s\leq1\}$$

Let $$c\in [0,1]$$

Define $$E_n=\{x|f_n(x)\geq cs(x)\}$$; $$n=1,2\ldots$$. Observe that $$E_1\subset E_2\subset\ldots$$ and $$\bigcup E_n=X$$

Suppose $$x\in X$$. If $$f(x)=0$$ then $$s(x)=0$$ implying that $$x\in E_1$$. If $$f(x)>0$$, then there exists $$n$$ such that $$f_n(x)>cs(x)$$ and hence, $$x\in E_n$$.

Thus, $$\bigcup E_n=X$$, therefore $$\displaystyle\int_Xf_n(x)d\mu\geq\int_{E_n}f_nd\mu\geq c\int_{E_n}sd\mu$$. As this is true if $$c\in [0,1]$$, we have that $$\alpha\geq\displaystyle\int_Xfd\mu$$. Thus, $$\displaystyle\int_Xf_nd\mu\to\int_Xfd\mu$$.

Fatou's Lemma
Let $$f_n\geq 0$$ be measurable functions. Then,

$$\displaystyle\int_X\liminf f_n d\mu\leq\liminf\int_Xf_nd\mu$$

Proof

For $$k=1,2,\ldots$$ define $$g_k(x)=\displaystyle\inf_{i\geq k}f_i(x)$$. Observe that $$g_k$$ are measurable and increasing for all $$x$$.

As $$k\to\infty$$, $$g_k(x)\to\displaystyle\liminf_{n\geq k}f_n(x)$$. By monotone convergence theorem,

$$\displaystyle\int_Xg_kd\mu\to\int_X\liminf f_n(x)d\mu$$ and as $$\displaystyle\int_Xg_k(x)d\mu\leq\liminf\int g_k(x)d\mu$$, we have the result.

Dominated convergence theorem
Let $$(X,\sigma,\mu)$$ be a complex measure space. Let $$\{f_n\}$$ be a sequence of complex measurable functions that converge pointwise to $$f$$; $$f(x)=\displaystyle\lim_{n\to\infty}f_n(x)$$, with $$x\in X$$

Suppose $$|f_n(x)|\leq g(x)$$ for some $$g\in\mathcal{L}^1(X)$$ then

$$f\in\mathcal{L}^1$$ and $$\displaystyle\int_X|f_n-f|d\mu\to 0$$ as $$n\to\infty$$

Proof

We know that $$|f|\leq g$$ and hence $$|f_n-f|\leq 2g$$, that is, $$0\leq 2g-|f_n-f|$$

Therefore, by Fatou's lemma, $$\displaystyle\int_X2gd\mu\leq\liminf\int_X(2g-|f_n-f|)d\mu\leq\displaystyle\int_X2gd\mu+\liminf\int_X(-|f_n-f|)d\mu$$

$$=\displaystyle\int_X2gd\mu-\limsup\int_X|f_n-f|d\mu$$

As $$g\in\mathcal{L}^1$$, $$\displaystyle\limsup_{n\to\infty}\int_X|f_n-f|d\mu\leq$$ implying that $$\displaystyle\limsup\int_X|f_n-f|d\mu$$

Theorem

 * 1) Suppose $$f:X\to[0,\infty]$$ is measurable, $$E\in\sigma$$ with $$\mu(E)>0$$ such that $$\displaystyle\int_Efd\mu=0$$. Then $$f(x)=0$$ almost everywhere $$E$$
 * 2) Let $$f\in\mathcal{L}^1(X)$$ and let $$\displaystyle\int_Efd\mu=0$$ for every $$E\in\sigma$$. Then, $$f=0$$ almost everywhere on $$X$$
 * 3) Let $$f\in\mathcal{L}^1(X)$$ and $$\displaystyle\left|\int_Xfd\mu\right|=\int_X|f|d\mu$$ then there exists constant $$\alpha$$ such that $$|f|=\alpha f$$ almost everywhere on $$E$$

Proof but $$\frac{1}{n}\mu(A_n)\leq\displaystyle\int_{A_n}fd\mu\leq\int_Efd\mu=0$$ Thus $$\mu(A_n)=0$$ for all $$n$$, by continuity, $$f=0$$ almost everywhere on $$E$$ Further as $$\displaystyle\int_Eu^+d\mu,\int_E(-u^-)d\mu$$ are both non-negative, each of them is zero. Thus, by applying part I, we have that $$u^+,u^-$$ vanish almost everywhere on $$E$$. We can similarly show that $$v^+,v^-$$ vanish almost everywhere on $$E$$.
 * 1) For each $$n\in\mathbb{N}$$ define $$A_n=\{x\in E|f(x)>\frac{1}{n}\}$$. Observe that $$A_n\uparrow E$$
 * 1) Write $$\displaystyle\int_Efd\mu=\int_Eu^+d\mu-\int_Eu^-d\mu+i\left(\int_Ev^+d\mu-\int_Ev^-d\mu\right)$$, where $$u^+,u^-,v^+,v^-$$ are non-negative real measurable.