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Suppose $Z$ is a random variable and $\epsilon>0$. I am trying to see when or if:

$$ \lim_{\epsilon \to 0}P(|Z-z| \leq \epsilon) = P(Z=z) $$

Is this a trivial result for $Z$ a discrete random variable? If $Z$ is continuous, must one use the fundamental theorem of calculus?

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It's possible to prove that $\lim_{\varepsilon\to0^+}\mathbb{P}(|Z-z|\leq \varepsilon)=\mathbb{P}(Z=z)$ just using properties of probability measures:

First note that if $0<\delta<\varepsilon$ then $$ \{Z=z\}\subset \{|Z-z|\leq \delta\}\subset \{|Z-z|\leq \varepsilon\}$$ so $$ \mathbb{P}(Z=z)\leq \mathbb{P}(|Z-z|\leq \delta)\leq \mathbb{P}(|Z-z|\leq \varepsilon)$$ Therefore it is enough to consider the limit as $\varepsilon\to 0$ through some sequence (say $\{\frac{1}{n}\}$), and since $$ \{Z=z\}=\bigcap_{n=1}^{\infty}\{|Z-z|\leq \frac{1}{n}\} $$ it follows from "continuity from above" that $$ \mathbb{P}(Z=z)=\lim_{n\to\infty}\mathbb{P}(|Z-z|\leq \frac{1}{n})=\lim_{\varepsilon\to0^+}\mathbb{P}(|Z-z|\leq \varepsilon)$$

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    Is this only for the case where $Z$ is discrete? – 2017-02-27
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    No, there is no hypothesis on $Z$ here (besides the fact that it is a random variable of course). – 2017-02-27
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    If $Z$ were to depend conditionally on some other random variable $X$, would it directly follow that $\lim_{\varepsilon\to0^+}\mathbb{P}(|Z-z|\leq \varepsilon | X)=\mathbb{P}(Z=z| X)$? Meaning, does the conditional version hold as well? – 2017-02-27
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    I think it should be possible to prove that using the dominated convergence theorem for conditional expectation. – 2017-02-27
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    More explicitly, are you referring to writing the conditional probability as the conditional expectation of an indicator random variable? – 2017-02-27
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    Yes, exactly. The idea would be to let $\varepsilon\to 0$ through a sequence again, and show that $\mathbb{E}[1_{|Z-z|\leq \varepsilon}\mid X]\to \mathbb{E}[1_{Z=z}\mid X]$. – 2017-02-27
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    I am having a bit of trouble showing this as the DCT requires $n \to \infty$ while we have $\epsilon \to 0$. I am trying to construct it by the set you have above, could I just redefine each of the events by putting in an $X$ as in: $\{Z=z, X\}\subset \{|Z-z|\leq \delta, X\}\subset \{|Z-z|\leq \varepsilon, X\}$ and then dividing the probability by $P(X)$ for the conditional? Thanks! – 2017-03-02
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    Now that I look at it, since conditional expectations are only unique up to sets of probability zero, you could potentially run into some issues involving uncountable unions of nullsets. – 2017-03-02
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    If $\mathbb{P}(\cdot\mid X)$ is a regular conditional probability (https://en.wikipedia.org/wiki/Regular_conditional_probability) then everything should be ok. – 2017-03-02