How is it possible for two random variables to have same distribution function but not same probability for every event? It is completely out of the world for me to hear that such a case exists. I was shocked and could not develop any intuition as to how it is possible. It also breaks my understanding (intuitive) of the distribution function. 
Here is the actual paragraph where this was given: 

 A: For your information, here is the relevant quote from 
page 27 of A Course in Probability Theory (2nd edition, Academic Press, 1974)
by Kai Lai Chung. He has just finished constructing the probability 
measure $\mu$ with distribution function $F$. 

There is one more question: besides the $\mu$ discussed above
  is there any other p.m. $\nu$ that corresponds to the given $F$ in the same
  way? It is important to realize that this question is not answered
  by the two preceding theorems. It is also worthwhile to remark that
  any p.m. $\nu$ that is defined on a domain strictly containing ${\cal B}^1$
   and that coincides with $\mu$ on ${\cal B}^1$ (such as the $\mu$ on ${\cal L}$
  as mentioned above) will certainly correspond to $F$ in the same way, and 
  strictly speaking such a $\nu$ is to be considered as distinct from $\mu$. 
  Hence we should phrase the question more precisely by considering only 
  p.m.'s on ${\cal B}^1$.

Here p.m. means probability measure, ${\cal B}^1$ are the Borel sets 
on the real line, and $\cal L$ is the $\mu$-completion of ${\cal B}^1$.
A: I'm going to stick my neck out and assert that the quotation is nonsense:

... it is possible for two random variables to have the same distribution function but not the same probability for every event.

It is nonsense because it does not make sense to say that a random variable "has a probability for an event". Once a probablity space is given, every event has a probability in itself (that's what the probability measure is for) -- and there's no role for a random variable in determining that probability.

What the quote appears to try to say is that $\mathbb R$ can be made into the sample space of a probability space in more than one way that gives the identity function (viewed as a random variable) the same distribution, but whose probability measures are not identical.
This can happen if the two probability measures are defined on a larger $\sigma$-algebra the algebra of Lebesgue-measurable sets, and assign different probabilities to some events that are not Lebesgue measurable.
Unfortunately, an example of this cannot be given explicitly; constructing one requires the Axiom of Choice. (Without AC it may be that all subsets or $\mathbb R$ are Lebesgue measurable and so their probabilities are given by the distribution).
In any case, it is confusing and misleading to describe this possibility as being about "two random variables" without further qualification, because that phrasing usually implies that the two random variables are defined on the same probability space (which makes the claim nonsense, as argued above).
A: Update: By Lebesgue-Stieltjes-Correspondence, there is a one to one correspondence between the probability law $P_X$ when $X$ is Borel measurable and the c.d.f $F_X$. I guess what the text meant is that we could have a random variable $Y$ being measurable to a richer sigma algebra, say, Lebesgue measurable sets than the Borel sets so that the correspondence breaks down.
Let $(\Omega,\mathcal{F},P)$ be the underlying probability space. Consider random variable $X$ and $Y$ defined on $(\Omega,\mathcal{F})$ with the following properties


*

*$X^{-1}(B)=Y^{-1}(B)\in \mathcal{F}$ for all Borel sets $B\in \mathcal{B}(\mathbb{R})$.

*$X^{-1}(C)\notin \mathcal{F}$ and $Y^{-1}(C)\in \mathcal{F}$, where $C$ is the famous subset of Cantor set that is Lebesgue measurable but not Borel measurable $C\in \mathcal{L}(\mathbb{R})\setminus \mathcal{B}(\mathbb{R})$.
This way we have a Borel-measurable but not Lebesgue measurable random variable $X$ and a Lebesgue measurable random variable $Y$. Recall that distribution function $F_X$ are induced by the law of $X$
$$
P(X\in B),\quad B\in \mathcal{B}(\mathbb{R}).
$$
It follows that $F_X=F_Y$ so $X$ and $Y$ have the same distribution functions.
However, $P(Y\in C)$ is a nonnegative real number whereas $P(X\in C)$ is not defined. Thus the probability (law) of the Lebesgue measurable but not Borel measurable set $C$ differs for $X$ and $Y$.
A tentative construction of such $X$ and $Y$:


*

*Let $(\Omega,\mathcal{F},P)$ be $([0,1],\mathcal{L}([0,1]),\lambda)$ where $\lambda$ is Lebesgue measure on $[0,1]$. Let the range of $X$ and $Y$ be $([0,1],\mathcal{L}([0,1]),\lambda)$ also.

*Let $\phi(x)$ be the Cantor function. Define $X$ by setting
$$
X^{-1}(x)=\phi(x)+x.
$$

*Let $C$ be the Cantor set and $\tilde{C}\subset C$ be the Lebesgue measurable but not Borel measurable set.

*By Vitali's Theorem, associate with $X^{-1}(C)$ a nonmeasurable subset $W\subset X^{-1}(C)$. The construction of $\tilde{C}$  (Royden, Real Analysis 4th Edition p.50) gives the following condition
$$
X(W)=\tilde{C},\quad X^{-1}(\tilde{C})=W\notin \mathcal{L}([0,1]).
$$

*Define the bijective transformation $S:\mathcal{L}([0,1])\cup\lbrace W\rbrace\rightarrow \mathcal{L}([0,1])\cup\lbrace W\rbrace$ such that
$$
S(X^{-1}(C))=W,\quad S(W)=X^{-1}(C),\quad \mbox{otherwise}.
$$

*Define $Y$ by setting
$$
Y^{-1}=S^{-1}X^{-1}.
$$

*Then for any $B\in \mathcal{B}([0,1])$, $B\neq W$ so $S^{-1}(B)=B$ and
$$
Y^{-1}(B)=S^{-1}X^{-1}(B)=X^{-1}(B).
$$
Property 1 is satisfied.

*For $\tilde{C}\in \mathcal{L}([0,1])\setminus \mathcal{B}([0,1])$, we have
$$
X^{-1}(\tilde{C})=W\notin \mathcal{L}([0,1])
$$
but
$$
Y^{-1}(\tilde{C})=S^{-1}X^{-1}(\tilde{C})=S^{-1}(W)=X^{-1}(C)\in\mathcal{L}([0,1]).
$$
Property 2 is satisfied.
