# Difference between $\lim_{n\to\infty}P(|X_n-X|<\epsilon)=1$ and $\lim_{n\to\infty}P(|X_n-X|=0)=1$

What is difference between $\lim_{n\to\infty}P(|X_n-X|<\epsilon)=1$ and $\lim_{n\to\infty}P(|X_n-X|=0)=1$? The former is the definition of convergence in probability. I'd like to get some intuitive understanding of the difference. My friend said the key to understanding the difference is that the former allows $\lim_{n\to\infty}P(0<|X_n-X|<\epsilon)>0$. Even if $\epsilon$ can be small arbitrarily, is this really possible?

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1. The thing is that even though $X_{n}\to X$, we might have $X_{n}(\omega)\neq X(\omega)$ for all $\omega\in\Omega$ and $n\in\mathbb{N}$. Since the sequence $(X_{n})$ converges to $X$, it is not necessarily the case that it ever reaches it. It just comes arbitrarily close to it. Hence $\lim_{n\to\infty}P(|X_{n}-X|=0)$ is not really a useful value to look at, as you take a limit of probabilities of $X_{n}$ being equal to $X$, which might never be the case for any $n\in\mathbb{N}$.

2. On the other hand, if you have $\lim_{n\to\infty}P(|X_{n}-X|<\varepsilon)=1$ for all $\varepsilon>0$, then this gives a sort of useful indication of $X_{n}$ being close to $X$. In the light of the previous analogy, if you have $X_{n}\to X$ so that $X_{n}$ never reaches $X$ but comes arbitrarily close to it, the probabilities of $X_{n}$ reaching any arbitrarily small fixed distance from $X$ is $1$.

Here is a link to another topic where I proved that $X_{n}\to X$ a.e. implies convergence with respect to the second notation, usually called "convergence in measure". Note that this implication requires $P(\Omega)<\infty$. E.g. a probability measure will do. I believe that in another topic you were looking for an answer to this question. Here is the proof: What does the following statement mean?

Note that in general the definition of convergence in measure is written in the complement form, because if $P(\Omega)=\infty$, then it still makes sense to talk about this concept. However if $P(\Omega)=1$, then these definitions coincide. By taking complements in the end you can modify the proof to suit your definition.

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Consider the probability space $([0,1],\mathcal{B}[0,1],\lambda|_{[0,1]})$ (where $\mathcal{B}$ denotes the borel-$\sigma$-algebra, $\lambda|_{[0,1]}$ the lebesgue measure on $[0,1]$) and

$$X_n(\omega) := \frac{1}{n} \qquad (\omega \in [0,1])$$

Then we know that $X_n \to 0=:X$ almost surely and $$\mathbb{P}[|X_n-X|=0] = \mathbb{P}[|X_n|=0] = 0$$

Let $\varepsilon>0$, then there exists $N \in \mathbb{N}$ such that $\frac{1}{N} \leq \varepsilon$. Hence for all $n \geq N$:

$$\mathbb{P}[|X_n-X|<\varepsilon] = \mathbb{P}[|X_n|<\varepsilon]=1$$

So we conclude:

$$\lim_{n \to \infty} \mathbb{P}[|X_n-X|=0] = 0 \qquad \lim_{n \to \infty} \mathbb{P}[|X_n-X|<\varepsilon]=1$$

which means that $\mathbb{P}[|X_n-X|<\varepsilon] \to 1$ does not imply $\mathbb{P}[|X_n-X|=0] \to 1$. The contrary is true since

$$1 \geq \mathbb{P}[|X_n-X| < \varepsilon] \geq \mathbb{P}[|X_n-X|=0] \to 1 \\ \Rightarrow \mathbb{P}[|X_n-X| < \varepsilon] \to 1$$

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