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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

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!

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    that looks weird. – 2017-01-21
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    @JorgeFernándezHidalgo why? – 2017-01-21
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    what is your definition of continuous variable? that the cummulative distribution function be continuous? – 2017-01-21
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    @JorgeFernándezHidalgo Is your question directed at me? If so, cdf can be written as $F_X(x)=\int_{-\infty}^{x}f_X(u)\text{ d}u$, for some non-negative function $f_X$. Therefore $F_X$ is continuous. – 2017-01-21
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    We can skip the integral part and just use that $F_X$ is continuous. Note that you can't use the fundamental theorem of calculus directly, because the only thing we know is that $\mu(X^{-1}(a,b])=F(b)-F(a)$. We can't do the same thing for closed intervals directly. – 2017-01-21
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    how? because the way I see it you don't know that $\mu(X^{-1}[a,b])=F(b)-F(a)$. – 2017-01-21
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    Sorry, I deleted my comment, because I'm not sure anymore if $F_X$ is continuous, because of the unbounded interval $(-\infty,x)$. I'm going to ask my teacher, or post a separate question on this forum. – 2017-01-21
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    The definition for continuous probability distribution is different in wikipedia. https://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution – 2017-01-21
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    Yea, I know. Using my definition, I can prove $\mathbb P(X=x)=0$, if $F_X$ is continuous. But I'll ask in a separate question how to prove the continuity of $F_X$. – 2017-01-21
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    We know, since $f_X$ is continuous, that it has an antiderivative, say $g$; then $\int_{x- \epsilon}^{x} f_X (u) \ du = g(x)-g(x- \epsilon)$, so $\lim_{\epsilon \downarrow 0} (g(x)-g(x- \epsilon)) = 0$. How do you know that the author of the proof didn't have that in mind as a given theorem when making that extra step? – 2017-01-21
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    @Pythagoricus First of all, how do we know $f_X$ is continuous? Second, that is exactly the step above... (line 2) – 2017-01-21
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    @Sha Vuklia I thought it is assumed to be... The step above is to say that the author might have gone back to the previous step by implicitly using the theorem about the existence of antiderivatives. – 2017-01-21
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    I added a proof that the function $F(x)=\int_{-\infty}^x fd\mu$ is a continuous function for every lebesgue-integrable function $f$. – 2017-01-21

2 Answers 2

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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 $F(x)=\nu(-\infty,x)$)

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    All of the technicalities are locked away inside the proof of those measure theory results really. – 2017-01-21
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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.

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    I'm sorry... That theorem does not help here... – 2017-01-21
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    Why not? We can fix $x$ and switch the interval endpoints, just like you said? – 2017-01-21
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    ...It says that $\chi \in [a,b]$, but we have $x$ as $a$ and then $c \uparrow x$, it can't happen. The interval should be $[x, b]$, but we need both $c \uparrow x$ and $c \in [x,b]$. – 2017-01-21
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    I think nonetheless that this is the theorem we are looking for... Perhaps with a modification of its proof such that it suits our case with the side limit. – 2017-01-21
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    Shouldn't the same theorem have an alternative statement like $F(\chi) = \int_{\chi}^{a} f(t) \operatorname{d}\!t$, $\chi \in [b,a]$? – 2017-01-21