# Measure-theoretic question about a relationship between the pdf and the distribution function of a random variable

For $$(\Omega, \Bbb F, \Bbb P)$$ a probability space and $$X: \Omega \to \Bbb R$$ a random variable, the distribution function of $$X$$ is defined as:

$$F_X(x) = \Bbb P(X^{-1}((-\infty,x]))$$

$$X$$ is called a continuous random variable if for some $$f_X \in L^1(\Bbb R)$$, such that $$f_X \ge 0$$, $$F_X (x) = \int_{-\infty}^x f_X(t)dt$$ for all $$x$$. In this case $$f_X$$ is called the pdf of $$X$$.

For a Borel set $$B$$, $$\Bbb P(X \in B) := \Bbb P(X^{-1}(B))$$.

If $$X$$ is a continuous random variable with pdf $$f_X$$, does the following hold for any Borel set $$B$$? Why?

$$\Bbb P(X \in B) = \int_B f_X(x)dx$$

This claim appears in a proof I am reading, and I don't understand why it holds. Thanks.

• First of all, you can understand this fact through the equivalence of measure and distribution function. Each distribution function defines a measure. So you can understand the integral as the Lebesgue one, and you are done. Any Borel set is from Borel sigma-algebra, so is Lebesgue measurable. Commented Nov 6, 2016 at 18:51
• @kolobokish I don't really understand what you are talking about. Would you suggest me some reference? Or can you explain your ideas with some more details in an answer? It doesn't have to answer the question.. anything helpful is welcome as an answer Commented Nov 6, 2016 at 19:00
• The answer really depends on your background knowledge and how you learned measure theory / integration (i.e. also which approach your lectures/textbook chose to use). You need to give some more details about what you know and what you have thought so far about this problem. This result is usually a trivial consequence of the definitions/theorems leading up to it.
– air
Commented Nov 6, 2016 at 19:28
• I mean, that every Borel set, is Lebesgue measurable. The integral of density function over the Borel set, is Lebesgue integral. The sigma- algebra of left interval of $(-\infty , a]$, is exactly the same sigma-algebra of of Borel sets. So any Borel set is in that sigma-algebra, and can be somehow obtained from that left-intervals. So the Lebesgue integral over Borel set can be obtained from the integrals with the measure you have defined for which you define you distributions. Or $\int_{B}fdP = \int_{R} f I_{B}dP$. $I_{B}$, can be represented somehow by the means of $I_{(-\infty, a]}$-s. Commented Nov 6, 2016 at 19:38

Yes, that statement is true. One way to rectify this definition with your definition of the probability density of the random variable $X$ is to start with open intervals. Suppose $B = (a,b)$. Then, express $\Bbb{P}(X\in B)$ using the distribution function as

$$\Bbb{P}(X\in B) = \int_Bf_X(x)dx = \int_a^b f_X(t)dt = F_X(b) - F_X(a)$$ Then, work up to a countable union of disjoint intervals. Say $B = \cup_{j=1}^\infty(a_j,b_j)$. By the countable additivity of measure,

$$\Bbb{P}(X\in \cup_{j=1}^\infty(a_j,b_j)) = \int_Bf_X(x)dx=\sum_{j=1}^\infty \int_{a_j}^{b_j}f_X(x)dx = \sum_{j=1}^\infty [F_X(b_j)-F_X(a_j)]$$ Then, for any Borel set $B$, you have to use the (measure theory) fact that $B$ can be approximated by a countable union of disjoint intervals. In fact, what you realize is that you are just constructing the definition of the Lebesgue integral of the function $\chi_B(x)f_X(x)\in L^1(\Bbb{R})$, where $\chi_B$ is the indicator function of $B$. In fact, once you convince yourself that the two definitions are equivalent, you can just take

$$\Bbb{P}(X\in B) :=\int_{-\infty}^\infty \chi_B(x)f_X(x)dx$$ as the definition of the probability measure for $X$.

• Thank you. I was trying to figure out what kind of approximation of $B$ by such a union will be useful here. Can you expand a little bit on that or give me some hint? Commented Nov 6, 2016 at 21:35
• It's too long of a story to put here. I would find a book on Lebesgue integration - "Measure Integral and Probability" by Capinski is probably at about the right level. Commented Nov 6, 2016 at 22:37

There is a connection between the abs. continuity of a random variable ($X$) and the abs. continuity of its distribution $P_X$ w.r.t. $m$ (the Lebesgue measure on $\mathbb{R}$). Namely, if $P_X$ is absolutely continuous w.r.t. $m$, then the Radon-Nikodym theorem implies that there exists $f_X=dP_X/dm$ (i.e. the R-N derivative of $P_X$ w.r.t. $m$), the pdf of $X$, such that for any Borel set $B$,

$$\mathbb{P}\{X\in B\}=P_X(B)=\int_B f_Xdm,$$

which is the result you're looking for.