Let ( X n ) n ∈ N ⊂ L 1 ( Ω , F , P ) (X_{n})_{n\in\mathbb{N}}\subset \mathscr{L}^{1}(\Omega,\mathcal{F},P) ( X n ) n ∈ N ⊂ L 1 ( Ω , F , P ) X n → X X_{n}\to X X n → X a.s. then X n → X in L 1 ⟺ ( X n ) n ∈ N uniformly integrable X_{n}\to X\text{ in }L^{1}\iff(X_{n})_{n\in\mathbb{N}}\text{ uniformly integrable} X n → X in L 1 ⟺ ( X n ) n ∈ N uniformly integrable L1 convergence is equivalent to being uniformly integrable .
\begin{proof} Fix ϵ > 0 \epsilon>0 ϵ > 0 Y n = 1 ∣ X n − X ∣ ≥ ϵ Y_{n}=\mathbb{1}_{|X_{n}-X|\ge \epsilon} Y n = 1 ∣ X n − X ∣ ≥ ϵ ∣ Y n ∣ ≤ 1 |Y_{n}|\le 1 ∣ Y n ∣ ≤ 1 lim n → ∞ Y n = 0 \lim_{ n \to \infty }Y_{n}=0 lim n → ∞ Y n = 0 a.s. (or E [ lim n → ∞ Y n ] = 0 \mathbb{E}[\lim_{ n \to \infty }Y_{n}]=0 E [ lim n → ∞ Y n ] = 0 Dominated Convergence Theorem E [ Y n ] → 0 \mathbb{E}[Y_{n}]\to0 E [ Y n ] → 0 P ( ∣ X n − X ∣ ≥ ϵ ) → 0 \mathbb{P}(|X_{n}-X|\ge\epsilon)\to0 P ( ∣ X n − X ∣ ≥ ϵ ) → 0 \end{proof}
If X n → X X_{n}\to X X n → X in probability , then there is a subsequence n k n_{k} n k X n k → X a.s. X_{n_{k}}\to X\,\text{a.s.} X n k → X a.s.
\begin{proof} P ( ∣ X n − X ∣ ≥ ϵ ) → 0 ⟹ ∃ X n k : P ( ∣ X n k − X ∣ ≥ 1 k ) ≤ 1 k 2 \mathbb{P}(|X_{n}-X|\ge \epsilon)\to0\implies \exists X_{n_{k}}:\mathbb{P}\left( |X_{n_{k}}-X|\ge \frac{1}{k} \right)\le \frac{1}{k^{2}} P ( ∣ X n − X ∣ ≥ ϵ ) → 0 ⟹ ∃ X n k : P ( ∣ X n k − X ∣ ≥ k 1 ) ≤ k 2 1 ∑ 1 k 2 < ∞ \sum \frac{1}{k^{2}}<\infty ∑ k 2 1 < ∞ Borel-Cantelli Lemma P ( { ∣ X n k − X ∣ ≥ 1 k i.o. } ) = 0. \mathbb{P}\left( \left\{ |X_{n_{k}}-X|\ge \frac{1}{k}\text{ i.o.} \right\} \right)=0. P ( { ∣ X n k − X ∣ ≥ k 1 i.o. } ) = 0. ∣ X n k − X ∣ < 1 k |X_{n_{k}}-X|< \frac{1}{k} ∣ X n k − X ∣ < k 1 k k k a.s. thus X n k → a . s . X . X_{n_{k}}\xrightarrow{a.s.} X. X n k a . s . X . \end{proof}
\begin{proof} P ( ∣ X n − X ∣ ≥ ϵ ) = P ( ∣ X n − X ∣ p ≥ ϵ p ) ≤ E [ ∣ X n − X ∣ p ] ϵ p → 0 \mathbb{P}(|X_{n}-X|\ge \epsilon)=\mathbb{P}(|X_{n}-X|^{p}\ge \epsilon^{p})\le \frac{\mathbb{E}[|X_{n}-X|^{p}]}{\epsilon^{p}}\to0 P ( ∣ X n − X ∣ ≥ ϵ ) = P ( ∣ X n − X ∣ p ≥ ϵ p ) ≤ ϵ p E [ ∣ X n − X ∣ p ] → 0 X n → L p X . X_{n}\xrightarrow{L^{p}}X. X n L p X . \end{proof}
If X n → X X_{n}\to X X n → X in probability and ∣ X n ∣ ≤ Y , ∀ n ∈ N |X_{n}|\le Y,\forall n\in\mathbb{N} ∣ X n ∣ ≤ Y , ∀ n ∈ N Y ∈ L p Y\in L^{p} Y ∈ L p X ∈ L p X \in L^{p} X ∈ L p X n → L p X X_{n}\xrightarrow{L^{p}}X X n L p X
>[!lemma] >If Z ∈ L 1 Z\in \mathscr{L}^{1} Z ∈ L 1 P ( A n ) → 0 \mathbb{P}(A_{n})\to0 P ( A n ) → 0 E [ Z ⋅ 1 A n ] = ∫ A n Z d P → 0 \mathbb{E}[Z\cdot \mathbb{1}_{A_{n}}]=\int\limits _{A_{n}}Z \, d\mathbb{P}\to0 E [ Z ⋅ 1 A n ] = A n ∫ Z d P → 0 \begin{proof} Step 1: X ∈ L p X\in \mathscr{L}^{p} X ∈ L p X n k → a . s . X X_{n_{k}}\xrightarrow{a.s.}X X n k a . s . X n k n_{k} n k ∣ X n k ∣ ≤ Y |X_{n_{k}}|\le Y ∣ X n k ∣ ≤ Y ∣ X ∣ ≤ Y |X|\le Y ∣ X ∣ ≤ Y a.s. .Step 2: X n → L p X X_{n}\xrightarrow{L^{p}}X X n L p X ϵ > 0 \epsilon>0 ϵ > 0 E [ ∣ X n − X ∣ p ] = E [ ∣ X n − X ∣ p ⋅ 1 { ∣ X n − X ∣ < ϵ } ] + E [ ∣ X n − X ∣ p ⋅ 1 { ∣ X n − X ∣ ≥ ϵ } ] ≤ ϵ p + E [ ( 2 Y ) p ⋅ 1 { ∣ X n − X ∣ ≥ ϵ } ] by lemma above → ϵ p ≡ 0 (since ϵ arbitrary) \begin{align*}
\mathbb{E}[|X_{n}-X|^{p}]&=\mathbb{E}[|X_{n}-X|^{p}\cdot \mathbb{1}_{\{ |X_{n}-X|<\epsilon \}}]+\mathbb{E}[|X_{n}-X|^{p}\cdot \mathbb{1}_{\{ |X_{n}-X|\ge\epsilon \}}]\\
&\le \epsilon^{p}+\mathbb{E}[(2Y)^{p}\cdot \mathbb{1}_{\{ |X_{n}-X|\ge \epsilon \}}]&\text{by lemma above}\\
&\to \epsilon^{p}\equiv0&\text{ (since }\epsilon \text{ arbitrary)}
\end{align*} E [ ∣ X n − X ∣ p ] = E [ ∣ X n − X ∣ p ⋅ 1 { ∣ X n − X ∣ < ϵ } ] + E [ ∣ X n − X ∣ p ⋅ 1 { ∣ X n − X ∣ ≥ ϵ } ] ≤ ϵ p + E [( 2 Y ) p ⋅ 1 { ∣ X n − X ∣ ≥ ϵ } ] → ϵ p ≡ 0 by lemma above (since ϵ arbitrary) \end{proof}
\begin{proof} By Portmanteau’s Theorem : X n → p X ⟹ P ( ∣ X n − X ∣ ≥ δ ) = 0 , ∀ δ > 0 X n → d X ⟹ E [ ψ ( X n ) ] → E [ ψ ( X ) ] , ∀ ψ ∈ C c 2 ( R ) \begin{gather*}
X_{n}\xrightarrow{p}X\implies \mathbb{P}(\left| X_{n}-X \right| \ge\delta)=0,\quad\forall\delta>0\\
X_{n}\xrightarrow{d}X\implies \mathbb{E}[\psi(X_{n})]\to \mathbb{E}[\psi(X)],\quad\forall\psi \in C_{c}^{2}(\mathbb{R})
\end{gather*} X n p X ⟹ P ( ∣ X n − X ∣ ≥ δ ) = 0 , ∀ δ > 0 X n d X ⟹ E [ ψ ( X n )] → E [ ψ ( X )] , ∀ ψ ∈ C c 2 ( R ) ∀ ψ ∈ C c 2 ( R ) , ∀ ϵ > 0 , ∃ δ > 0 : ∣ X − Y ∣ ≤ δ ⟹ ψ ( X ) − ψ ( Y ) < ϵ \forall\psi \in C_{c}^{2}(\mathbb{R}),\forall\epsilon>0,\exists\delta>0:|X-Y|\le\delta\implies\psi(X)-\psi(Y)<\epsilon ∀ ψ ∈ C c 2 ( R ) , ∀ ϵ > 0 , ∃ δ > 0 : ∣ X − Y ∣ ≤ δ ⟹ ψ ( X ) − ψ ( Y ) < ϵ which holds by compact ness. Thus, if M = sup x ∣ ψ ( x ) ∣ M=\sup_{x}|\psi(x)| M = sup x ∣ ψ ( x ) ∣ ϵ > 0 \epsilon>0 ϵ > 0 ∣ E [ ψ ( X n ) ] − E [ ψ ( X ) ] ∣ ≤ E [ ∣ ψ ( X n ) − ψ ( X ) ∣ ] triangle inequality = E [ ∣ ψ ( X n ) − ψ ( X ) ∣ ⋅ 1 ∣ X n − X ∣ ≤ δ ] + E [ ∣ ψ ( X n ) − ψ ( X ) ∣ ⋅ 1 ∣ X n − X ∣ > δ ] ≤ ϵ + 2 M ⋅ P ( ∣ X n − X ∣ > δ ) → ϵ
\begin{align*}
\left| \mathbb{E}[\psi(X_{n})]-\mathbb{E}[\psi(X)] \right| &\le \mathbb{E}[\left| \psi(X_{n})-\psi(X) \right| ]&\text{triangle inequality}\\
&= \mathbb{E}[\left| \psi(X_{n})-\psi(X) \right| \cdot \mathbb{1}_{|X_{n}-X|\le\delta}]\\
&\quad\quad\quad\quad +\mathbb{E}[\left| \psi(X_{n})-\psi(X) \right| \cdot \mathbb{1}_{|X_{n}-X|>\delta}]\\
&\le \epsilon+2M\cdot \mathbb{P}(|X_{n}-X|>\delta)\\
&\to\epsilon
\end{align*}
∣ E [ ψ ( X n )] − E [ ψ ( X )] ∣ ≤ E [ ∣ ψ ( X n ) − ψ ( X ) ∣ ] = E [ ∣ ψ ( X n ) − ψ ( X ) ∣ ⋅ 1 ∣ X n − X ∣ ≤ δ ] + E [ ∣ ψ ( X n ) − ψ ( X ) ∣ ⋅ 1 ∣ X n − X ∣ > δ ] ≤ ϵ + 2 M ⋅ P ( ∣ X n − X ∣ > δ ) → ϵ triangle inequality \end{proof}
So, we have the following picture:
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