Let random variables$X_1,X_2,\cdots,X_n$ be i.i.d. and any $E(X_i)=\mu<\infty$. If we put
\[ Y_n=\frac{1}{n}\sum_{i=1}^nX_i \]
then law of large numbers says,
\[ P(\lim_{n\rightarrow\infty}Y_n=\mu)=1 \]
It means, for any $\epsilon>0$ there is a $N$ such that if $n>N$, $|Y_n-\mu|\leq \epsilon$ and it's probability becomes 1.
In the probability theory, as we are used to write elements $w$ of the sample space explicitly,
\[ P (w |\lim_{n\rightarrow\infty}Y_n(w)=\mu )=1. \]
Therefore, when $1/j,(j\in \mathbb{N})$ is used in place of $\epsilon$, if a set $A$ is defined by
\[ A=\left\{w |\forall j, \exists N, n\geq N, |Y_n-\mu|<1/j \right\} \]
, then $P(A)=1$ . Furthermore, using symbols of the set theory, if a set $A$ is
\[ A=\bigcap_{j=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n=N}^{\infty} \left\{w | |Y_n-\mu|<1/j \right\} \]
, then $P(A)=1$ .
A epsilon-delta technique can be written in the form like this.
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