• Monotone converge theorem
  • Construct measure from integrals
  • Radon-Nykodym Theorem
  • Uniform integrability
• Fatou’s Lemma
• Dominated convergence theorem

In this section, we talk about the the theorems related to the convergence of sequence of integrals.

Monotone converge theorem

THEOREM Suppose $f_n\geq 0$ and $f_n\uparrow f$, where $f_n\in \mathcal{F}$ on $(\Omega,\mathcal{F},\mu)$, then $\int f_nd\mu\uparrow \int f d\mu$.

PROOF Suppose for any fixed $n$, there exists a simple function sequence $g_{n,k}\uparrow f_n$,

then $\int f_nd\mu = \lim_k \int g_{n,k}d\mu$, $\forall n$.

Let $g_k=\max_{1\leq n\leq k}g_{n,k}$, we prove that $\{g_k\}$ is a sequence of simple functions converge to $f$.

1. $g_k\leq g_{k+1}$
2. $g_k\leq f_k$
3. $g_k\uparrow f$

Now we have $\int fd\mu = \lim_k \int g_kd\mu$.

$\int f_nd\mu\leq \int fd\mu \Longrightarrow \lim\int f_nd\mu\leq \int fd\mu$. We have to prove $\lim\int f_nd\mu\geq \int fd\mu$. $g_k\leq f_k\Longrightarrow \int f d\mu = \lim_k \int g_kd\mu \leq \lim_k \int f_kd\mu$. $\square$

REMARK if the theorem is $f_n\geq g$, $g$ integrable, then we can consider the sequence $g_n=f_n-f$, and the result still holds.

Construct measure from integrals

Suppose $f\geq 0$, $\mu_f(A)=\int_Afd\mu$, then

1. $\mu_f(\emptyset)=0$
2. $\mu_f$ is $\sigma$-additive.

OBS Suppose $B\in \mathcal{F}$ and $\mu(B)=0$, then $\int_B fd\mu=0$, $\mu_f(B)=0$.

DEF (Absolute continuity) Suppose $\mu, \nu$ on $(\Omega, \mathcal{F})$, $\mu$ is absolute continuous with respect to $\nu$ ($\mu\ll \nu$) if $\nu(A)=0\Longrightarrow \mu(A)=0$.

Radon-Nykodym Theorem

THM If $\mu \ll \nu$, $\exists g=\frac{d\mu}{d\nu}\geq0$ s.t. $\mu(B)=\int_Bgd\nu$.

It is possible for two measures $\mu, \nu$ that neither $\mu\ll \nu$ nor $\nu \ll \mu$. Assume $f\geq 0$ on $\mathbb{R}$, $\lambda$ to be the Lebesgue measure, $\mu_f(A)=\int_Afd\lambda$,

$\forall x \in B$, $\mu(\left{x\right})=0$, $\delta_x(\left{x\right})=1$, thus $\delta_x\ll \mu_f$ is false. $\forall [a,b)\notin \mathbb{R}$, $\delta_x([a,b))=0$, $\mu_f([a,b))>0$, thus $\mu_f \ll \delta_x$ is false.

Uniform integrability

$\int fd\mu\leq \infty$, $\forall \varepsilon>0$, $\exist \delta > 0$, $\mu(A)<\delta\Longrightarrow \mu_f(A)<\varepsilon$.

Fatou’s Lemma

If $f_n\geq g$, where $g$ is integrable, $\int \underline{\lim}f_nd\mu\leq \underline{\lim}\int f_nd\mu$.

If $f_n\leq g$, where $g$ is integrable, $\int \overline{\lim}f_nd\mu\geq \overline{\lim}\int f_nd\mu$.

Dominated convergence theorem

PROOF

Thus all “$\geq$” are “$=$”.