Beppo Levi Theorem

Theorem (1.27)

If fn:X→[0,∞]f_{n}:X\to[0,\infty] is measurable, for n=1,2,3,…,n=1,2,3,\dots, and f(x)=∑n=1∞fn(x)x∈Xf(x)=\sum_{n=1}^{\infty}f_{n}(x)\quad x \in Xthen∫Xf dμ=∑n=1∞∫Xfn dμ\int\limits _{X}f \, d\mu =\sum_{n=1}^{\infty}\int\limits _{X}f_{n} \, d\mu which also shows that ∫(f1+f2) dμ=∫f1 dμ+∫f2 dμ\int\limits (f_{1}+f_{2}) \, d\mu=\int\limits f_{1} \, d\mu+\int\limits f_{2} \, d\mu

Theorem (1.38)

Let (fn)n∈N(f_{n})_{n\in\mathbb{N}} be a sequence of C\mathbb{C}-Measurable Functions defined a.e. on XX s.t. ∑n=1∞∫X∣fn∣ dμ<∞\sum_{n=1}^{\infty}\int\limits _{X}|f_{n}| \, d\mu<\infty Then the series f(x)=∑n=1∞fn(x)f(x)=\sum_{n=1}^{\infty}f_{n}(x)converges almost for all x,f∈L1(μ)x,f\in L^{1}(\mu) and ∫Xf dμ=∑n=1∞∫Xfn dμ\int\limits _{X}f \, d\mu=\sum_{n=1}^{\infty}\int\limits _{X}f_{n} \, d\mu

Lemma (Beppo-Levi (Probability))

For rv Y1,Y2,⋯∈L1Y_{1},Y_{2},\dots \in \mathscr{L}^{1} with Yi≥0,∀i≥1Y_{i}\ge 0,\forall i\ge 1, if ∑n=1∞E[Yn]=c<∞\sum_{n=1}^{\infty}\mathbb{E}[Y_{n}]=c<\infty then

  1. ∑n=1∞Yn<∞\sum_{n=1}^{\infty}Y_{n}<\inftya.s. and consequently;
  2. Yn→a.s.0Y_{n}\xrightarrow{a.s.}0

\begin{proof} Let EE be the event {∑n=1∞Yn=∞}\left\{ \sum_{n=1}^{\infty}Y_{n}=\infty \right\}. For a large constant T>0T>0, define the rv SN=min⁡{T,∑n=1NYn}S_{N}=\min\left\{ T,\sum_{n=1}^{N} Y_{n} \right\}SNS_{N} is a bounded rv, SNS_{N} is non-decreasing, so, it has a limit and for ω∈E\omega \in E, lim⁡N→∞SN(ω)=T\lim_{ N \to \infty } S_{N}(\omega)=TBy Dominated Convergence Theorem we have that lim⁡N→∞E[SN]=E[lim⁡N→∞SN]=T≥T⋅P(E)\lim_{ N \to \infty } \mathbb{E}[S_{N}]=\mathbb{E}\left[ \lim_{ N \to \infty } S_{N} \right]=T\ge T\cdot \mathbb{P}(E)But, lim⁡N→∞E[SN]≤lim⁡N→∞E[∑n=1NYn]=E[lim⁡N→∞∑n=1NYn]=c\begin{align*} \lim_{ N \to \infty }\mathbb{E}[S_{N}]&\le \lim_{ N \to \infty } \mathbb{E}\left[ \sum_{n=1}^{N}Y_{n} \right] \\ &= \mathbb{E}\left[ \lim_{ N \to \infty } \sum_{n=1}^{N}Y_{n} \right]\\ &= c \end{align*}Hence, P(E)≤cT\mathbb{P}(E)\le \frac{c}{T}. But TT is arbitrary so T→∞T\to \infty yields P(E)=0\mathbb{P}(E)=0.

\end{proof}

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