Specific step in proof $\mathbb P(X=x)=0$ I have a question about the following proof, where it's shown that $\mathbb P(X=x)=0$, for a continuous variable $X$ with dentisty function $f_X$:
\begin{align}
\mathbb P(X=x)=&\lim_{\epsilon\downarrow 0}\mathbb P(x-\epsilon<X\leq x)\\
=&\lim_{\epsilon\downarrow 0}\left[F_X(x)-F_X(x-\epsilon)\right]\\
=& \lim_{\epsilon\downarrow0}\int_{x-\epsilon}^{x}f_X(u)\text{ d}u\\
=&\ 0.
\end{align}
First of all, why didn't the writer stop at the second line. By the fundamental theorem of calculus, we know that $F_X$ is a continuous, so $\lim_{\epsilon\downarrow 0} F_X(x-\epsilon)= \lim_{t\to x}F_X(t)=F_X(x)$. 
But that's not my biggest issue. Assume we would really like to conclude from the integral bit that $\mathbb P(X=x)=0$; which theorems to use?
So I'm specifically asking about this step: $\lim_{\epsilon\downarrow0}\int_{x-\epsilon}^{x}f_X(u)\text{ d}u=0$.
Thanks in advance!
 A: According to this wikipedia page a continuous variable is one such that its cummulative distribution fucntion is continuous.
Using this we can notice that $F(x)-F(x-\epsilon)=\mu(X^{-1}(x-\epsilon,x])\geq \mu(X^{-1}\{x\})$ for all $\epsilon$, and because $F$ is continuous the left side goes to $0$ when $\epsilon$ goes to zero. We conclude $\mu(X^{-1}\{x\})=0$ as desired.

In general, if $\nu$ is any measure and you have a family of measurable sets $A_1\subseteq A_2\dots $ such that their union is $A$ you have $\lim_{n\to \infty} \nu(A_n)=A$, and if you have measurable sets $B_1\supseteq B_2 \supseteq \dots$ such that their intersection is $B$ then you have $\lim_{n\to \infty} \nu(B_n)=B$. This result is sometimes called convergence of measures.
Notice that if $f$ is an integrable function then $\nu(A)=\int_A f d\mu$ is a measure. It follows that $\lim_{n\to \infty} \nu(-\infty,x+1/n]=\nu(-\infty,x)$ and $\lim_{n\to \infty} \nu(-\infty,x-1/n)=\nu(-\infty,x]$.
it follows that $F$ is continuous (because $F(x)=\nu(-\infty,x)$)
A: For the last step of the proof in question:
$$\lim_{\epsilon \downarrow 0} \int_{x- \epsilon}^{x} f(x) \operatorname{d}\!x = \lim_{c \uparrow x} \int_{c}^{x} f_X(u) \operatorname{d}\!u = \int_{x}^{x} f_X(u) \operatorname{d}\!u = 0,$$
where we used the the theorem that states that the integral function
$$F(\chi) = \int_{\chi}^{a} f(t) \operatorname{d}\!t, \quad \chi \in [b,a],$$
is (uniformly) continuous. Notice that $x$ is taken as a parametre in the original function.
P.S.: I've checked the proof of the above mentioned theorem from a book; $F$ is usually defined as
$$F(\chi) = \int_{a}^{\chi} f(t) \operatorname{d}\!t, \quad \chi \in [a,b],$$
in standard statements of the theorem, but what I write here is equally valid since the same proof applies to it! [I'll update this answer tomorrow to include that proof.]

UPDATE:
We will prove that if $f$ is an integrable function over $[b,a]$, then function
$$F(\chi) = \int_{\chi}^{a} f(t) \operatorname{d}\!t, \quad \chi \in [b,a],$$
is uniformly continuous.
Let $\chi, \psi \in [b,a]$; then
$$F(\psi) - F(\chi) = \int_{\chi}^{\psi} f(t) \operatorname{d}\!t.$$
If $\chi \leq \psi$, then, by the triangle inequality for integrals, we have
$$\left | \int_{\chi}^{\psi} f(t) \operatorname{d}\!t \right | \leq \int_{\chi}^{\psi} | f(t) | \operatorname{d}\!t;$$
similarly, if $\chi > \psi$ we get
$$\left | \int_{\chi}^{\psi} f(t) \operatorname{d}\!t \right | \leq \int_{\psi}^{\chi} | f(t) | \operatorname{d}\!t.$$
So, in either case it holds that
$$| F(\chi) - F(\psi) | \leq \left | \int_{\chi}^{\psi} | f(t) | \operatorname{d}\!t \right | \leq \sup_{b \leq t \leq a} |f(t)| \ |\chi - \psi|.$$
Now, fix a positive $\epsilon$. There exists a positive $\delta = \frac{\epsilon}{\sup_{b \leq t \leq a} |f(t)|}$ such that
$$|\chi - \psi|< \delta \implies |F(\chi) - F(\psi)| < \epsilon$$
(for all $\chi, \psi \in [b,a]$). This means that $F$ is uniformely continuous, and the proof is complete.
The exact same proof applies even if we define $F$ as
$$F(\chi) = \int_{a}^{\chi} f(t) \operatorname{d}\!t, \quad \chi \in [a,b].$$

UPDATE 2: On the continuity of the Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is defined as
$$F_X(x) = \int_{-\infty}^{x} f_X(t) \operatorname{d}\!t, \quad x \in \mathbb{R}.$$
We wish to prove that this function is continuous at any $a \in \mathbb{R}$, so pick an $a$ and then a constant $c < a$; now you can write
$$F_X(x) = \int_{-\infty}^{c} f_X(t) \operatorname{d}\!t + \int_{c}^{x} f(t) \operatorname{d}t$$
for all $x \geq c$. By what was shown above, the second integral is continuous at $x=a$ and so is $F_X(x)$!
If you want to know more about functions like the CDF you can have a look at Wikipedia https://en.wikipedia.org/wiki/Absolute_continuity. Notice that what all integrals appearing in my post are assumed to be Riemann, or generalised Riemann, integrals.
