In the Introduction to Probability course of MIT (6.012), we are assuming that twelve equiprobable elementary events are taking place in a universe $\Omega$, as indicated in this Venn diagram: 
Professor then says "since the twelve points are equiprobable in $\Omega$, when assuming $B$ has happened, the remaining six points in $B$ must be equiprobable as well".
I don't know much about probability theory (even though I am "reading" Jaynes' book Probability Theory, but I skip a lot in the book) neither do I know about measure theory but I was wondering if there could be an argument that shows equiprobability can be extended to the elements of $B$ (when considering that $B$ has happened) ?
Thanks
I'm using the term "equiprobable" in my answer below. That term is certainly correct and appropriate, but often, especially in the continuous setting, you'll probably hear "uniform" instead of "equiprobable."
For a probability space $(\Omega,\mathcal{B},\mathbb{P})$ and $B\in\mathcal{B}$ with $\mathbb{P}(B)>0$, the conditional probability of $A\in\mathcal{B}$ given that $B$ has occurred is (by usually by definition) $$\mathbb{P}(A|B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}\tag{1}$$ So, in the Venn diagram, we're restricting to subsets of $B$ (that is why we have $A\cap B$, so we are only concerned with the part of $A$ that is in $B$) and dividing by $\mathbb{P}$ (so that the conditional probability of $B$ given that $B$ has occurred is $\mathbb{P}(B\cap B)/\mathbb{P}(B)=1$).
For a finite probability space $(\Omega,\mathcal{B},\mathbb{P})$, "equiprobable" means that there exists $p$ such that, if $x_1,\ldots,x_N$ are the members of the probability space, $\mathbb{P}(\{x_i\})=p$ for each $i=1,2,\ldots,p$. Then $$1=\mathbb{P}(\Omega)=\sum_{i=1}^N \mathbb{P}(\{x_i\})=pN,$$ and $p=1/N$. So, in other words, each individual outcome $x_i$ satisfies $\mathbb{P}(\{x_i\})=1/N$. Then if $A\subset \Omega$ is a non-empty subset of $\Omega$ and if $x_{i_1},\ldots,x_{i_{|A|}}$ are the members of $A$, then $$\mathbb{P}(A)=\sum_{j=1}^{|A|}\mathbb{P}(\{x_{i_j}\})=|A|/N.$$ This is clearly also true if $A=\varnothing$, since $\mathbb{P}(\varnothing)=0=0/N$. So we have $\mathbb{P}(A)=|A|/N=|A|/|\Omega|$ for $A\subset \Omega$ in the case of equiprobability. The converse is also true. That is, if $(\Omega,\mathcal{B},\mathbb{P})$ is a finite probability space with $\mathbb{P}(A)=|A|/|\Omega|$ for all $A\subset \Omega$, then $(\Omega,\mathcal{B},\mathbb{P})$ is an equiprobable space. To see this, we can just take all sets $A$ with $|A|=1$. Now assume $(\Omega,\mathcal{B},\mathbb{P})$ is a finite equiprobable probability space and fix $\varnothing\neq B\subset \Omega$. Then for any $A\subset B$, $$\mathbb{P}(A|B)=\frac{\mathbb{P}(A)}{\mathbb{P}(B)}=\frac{|A|/|\Omega|}{|B|/|\Omega}=\frac{|A|}{|B|}.$$ In particular, for $x\in B$, $\mathbb{P}(\{x\}|B)=1/|B|$. This shows that $B$ is equiprobable.
The preceding deals with discrete equiprobable spaces. On the other hand, in the continuous setting, "equiprobable" means that we have a probability density function which is constant everywhere that it is positive. That is, suppose that $f:\mathbb{R}\to\mathbb{R}$ and $\Omega\subset \mathbb{R}$ are such that $$f(x)=\left\{\begin{array}{ll} c & : x\in \Omega \\ 0 & : x\in \mathbb{R}\setminus \Omega,\end{array}\right.$$ from which it follows that and $$\mathbb{P}(X\in A)=\int_A f(x)dx = \int_{A\cap \Omega} c dx = cm(A),$$ where $m(A)$ is the measure of $A$. We have $$1=\int_{-\infty}^\infty f(x)dx=\int_\Omega c dx = cm(\Omega),$$ so $c=1/m(\Omega)$ (which is directly analogous to $1/N$ in the discrete case). Thus the pdf of a continuous, equiprobable distribution on $\Omega\subset \mathbb{R}$ is $f(x)=\frac{1_\Omega}{m(\Omega)}$, where $1_\Omega$ is the indicator of $\Omega$ ($1_\Omega(x)=1$ if $x\in \Omega$ and $1_\Omega(x)=0$ otherwise). So $$\mathbb{P}(X\in A)=\int_A \frac{1_\Omega(x)}{m(\Omega)}dx$$ in this case.
For $B\subset \Omega$ and $A\subset B$, $$\mathbb{P}(X\in A|B)=\frac{\int_A f(x)dx}{\int_B f(x)dx} = \frac{m(A)/m(\Omega)}{m(B)/m(\Omega)}=\frac{m(A)}{m(B)}=\int_A \frac{1_B}{m(B)}(x)dx.$$ So the pdf of $\mathbb{P}(\cdot|B)$ is $\frac{1_B}{m(B)}$, which is constant on $B$ and $0$ otherwise. Again, this is the equiprobable distribution on $B$.