Measure Theory/Integration

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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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

  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

  1. For each n\in\mathbb{N} define A_n=\{x\in E|f(x)>\frac{1}{n}\}. Observe that A_n\uparrow E
    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
  2. 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.
    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.