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I am wondering whether the following result is true:

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of i.i.d. real-valued random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with law $\Lambda$. Then $$ \limsup_{n \to \infty} \mathbb{P} \bigg[ \frac{1}{n} \sum_{i=1}^n \mathbf{1}_{A} (X_i) =\Lambda (A), \quad \forall A \in \mathcal{B} ( \mathbb{R}) \bigg] =1.$$

By the strong law of large numbers, it is clear that $$ \frac{1}{n} \sum_{i=1}^n \mathbf{1}_{A} (X_i) \xrightarrow{n \to \infty} \Lambda (A), \quad \forall A \in \mathcal{B} ( \mathbb{R}), \quad \text{a.s.}$$ However, it is not obvious whether this statement is true or not. I tried applying Fatou's lemma but that doesn't seem to help. Any ideas?

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closed as unclear what you're asking by Did, Leucippus, R_D, choco_addicted, user223391 Jun 24 '16 at 22:27

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Certainly not. Unless the $X_i$ are constant, the probability that $\frac{1}{n} \sum \limits_{i = 1}^n 1_A(X_i)$ equals $\Lambda(A)$ for all $A \in \mathcal{B}(R)$ is $0$ for any $n$. You need to rethink what you want to show.

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