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About the first Borel-Cantelli lemma


The first BorelCantelli lemma is one of rare elementary providers of almost sure convergence in probability theory. It states that if \( {(A_n)} \) is a sequence of events in a probability space \( {(\Omega,\mathcal{F},\mathbb{P})} \) such that \( {\sum_n\mathbb{P}(A_n)<\infty} \) then \( {\mathbb{P}(\varlimsup_nA_n)=0} \). The classical proof consists in noticing that \( {B_n=\cup_{m\geq n}A_m} \) is non increasing in \( {n} \), and thus, since \( {\sum_n\mathbb{P}(A_n)<\infty} \),

\[ \mathbb{P}(\varlimsup A_n) =\mathbb{P}(\cap_nB_n) =\lim_n\mathbb{P}(B_n) \leq \lim_{n}\sum_{m\geq n}\mathbb{P}(A_m)=0. \]

There is another proof of this lemma that I prefer, and which goes as follows: by the Fubini-Tonelli theorem or by the monotone convergence theorem,

\[ \mathbb{E}(\sum_n\mathbf{1}_{A_n}) =\sum_n\mathbb{E}(\mathbf{1}_{A_n}) =\sum_n\mathbb{P}(A_n) <\infty \]

therefore \( {\mathbb{P}(\sum_n\mathbf{1}_{A_n}=\infty)=0} \). Now \( {\{\sum_n\mathbf{1}_{A_n}=\infty\}=\varlimsup A_n} \). I like very much this proof because it concentrates useful ingredients for students:

  • we count by summing indicators of events and \( {\{\sum_n\mathbf{1}_{A_n}\}=\varlimsup A_n} \)
  • \( {\mathbb{E}(\mathbf{1}_A)=\mathbb{P}(A)} \)
  • Fubini-Tonelli allows to swap \( {\mathbb{E}} \) and \( {\sum} \)
  • if \( {X\geq0} \) and \( {\mathbb{E}(X)<\infty} \) then \( {\mathbb{P}(X=\infty)=0} \)

A very similar argument may be used to obtain quickly a strong law of large numbers. Namely, if \( {(X_n)} \) are independent centered random variables bounded in \( {\mathbf{L}^4} \) (i.e. \( {\sup_n\mathbb{E}(|X_n|^4)<\infty} \)) and if we set \( {S_n:=\frac{1}{n}(X_1+\cdots+X_n)} \), then by expanding and using the assumptions we get \( {\mathbb{E}(S_n^4)=\mathcal{O}(n^{-2})} \), which gives

\[ \mathbb{E}(\sum_nS_n^4) =\sum_n\mathbb{E}(S_n^4) <\infty \]

and thus \( {\mathbb{P}(S_n\rightarrow0)\geq\mathbb{P}(\sum_nS_n^4<\infty)=1} \). To me, this is a bit more elegant than using the first Borel-Cantelli lemma and the Markov inequality.

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