Limit of probabilities of a large sample

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As the sample size n approaches infinity, the probability of the sample mean converging to the population mean is asserted to be 1, supported by the Central Limit Theorem. The sample mean distribution approximates a Normal distribution with specified parameters, leading to a delta function centered at the population mean. However, since the sample mean is a continuous random variable, the probability of it equaling any specific value, including the population mean, remains zero. This indicates that while the sample mean approaches the population mean, the probability of exact equality does not increase. The discussion highlights the nuanced understanding of convergence in probability theory.
MAXIM LI
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Homework Statement
## Let {X_n}_{n≥1}## be a sequence of iid random variables having a common density function
## f(x) = \begin{cases} xe^{-x} &\text{ for } x \ge 0 \\ 0 &\text{ otherwise }\end{cases}##

Let ##\bar{X}_n = \frac{1}{n}\sum_{i=1}^{n} X_i## where ##n=1,2,\ldots##. Then find ##\lim_{{n\to\infty}} P(\bar{X}_n=2)##
Relevant Equations
##\lim_{{n\to\infty}} P(\bar{X}_n=2)##
My first thought as well but I think the problem is deeper than that. I think that as the n tends towards infinity the probability of the the sample mean converging to the population mean is 1. Looking at proving this.
By the Central Limit Theorem the sample mean distribution can be approximated by a Normal distribution with $$\mu = 2,~\sigma = \sqrt{\dfrac{2}{n}}$$

As ##n\to \infty## this becomes a delta function centered at ##2##
 
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You're overcomplicating this. ##\overline{X}_n## is a continuous random variable so ##P(\overline{X}_n = a) = 0## for all ##a \in \mathbb{R}##. In particular
$$\lim_{n \to \infty} P(\overline{X}_n = 2)= \lim_{n \to \infty} 0 = 0$$
 
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First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...