Probability measures, Integral I had a test and was asked to prove that, with $\Omega = (0,1)$,
$$
\mathbb{P}(B)= \frac32 \int_B \sqrt{x} \,dx,
$$
is a probability measure, meaning it has to satisfy that $\mathbb{P}(\Omega)=1$, but had the condition $0 < x < 1$ (i.e. $B \subseteq \Omega$).
I agree that
$$\mathbb{P}(\Omega) = \left. x^{3/2}\right|_0^1 = 1,$$
but $x$ can’t take those values, so $\mathbb{P}(\Omega)$ cannot be equal to 1, which means that $\mathbb{P}(B)$ is not a probability measure.
But my teacher insists that it is an integral, so taking away those points doesn’t affects the area, I don’t know which is the right answer because I don’t understand that argument, because an area is a group of “dots” joined together.
 A: Your teacher is right.  $\frac{3}{2}\int_\Omega \sqrt{x}\;dx = 1$ where $\Omega = (0,1)$ is the open interval.
I do not agree with: 

an area is a group of “dots” joined together

A: Informally speaking, you define the definite integral $$\int_B f(x) dx$$ to be the area under the curve of $f(x)$ over $B \subseteq \mathbb{R}$. Just like length of the interval $(0,1)$ is the same as the interval $[0,1]$, and is unaffected by adding or removing a countable amount of points, so too area is unaffected by adding or removing any countable amount of lengths from it.

Formally speaking, for any interval $A = [a_-, a_+]$ you define
$$
\int_A f(x) dx = \lim_{n \to \infty} \sum_{k=1}^n f(x_k) \Delta x_k
$$
where typically you assume a uniform partition of $A$ so $$\Delta x_k = (a_+ - a_-)/n$$ and $$k-1 < \frac{x_k - a_-}{\Delta x} < k, \quad \forall k.$$
It can be shown further that you can integrate over any partition of $A$ you like, not just a uniform one, and then you will have $\Delta x_k$ actually depend on $k$. Either way, the critical idea is for you to still be able to pick a point $x_k$ inside each element of the partition, which can be done as long as you have your function defined everywhere except a countable number of points of $A$.

Finally, your teacher is correct, and $\mathbb{P}[\cdot]$ as defined above is indeed a valid distribution.
A: $\def\e{\varepsilon}$The (Riemann) integral of a suitable well-behaved function $f(x)$ is defined over any closed interval $[a,b]$. To extend this definition to open and unbounded intervals, the integral over an open interval $(a,b)$ is defined as a limit:
$$
\int_{(a,b)} f(x)\,dx\stackrel{\text{def}}=\lim_{\e_1\to 0^+,\e_2\to 0^+} \int_{a+\e_1}^{b-\e_2}f(x)\,dx
$$
This is referred to as an improper integral.
In your case,
$$
\frac32\int_{(0,1)}\sqrt{x}\,dx=\lim_{\e_1\to 0^+,\e_2\to 0^+} \frac{3}2\int_{\e_1}^{1-\e_2}\sqrt{x}\,dx=\lim_{\e_1\to 0^+,\e_2\to 0^+}(1-\e_2)^{3/2}-\e_1^{3/2}=1
$$
so it works as you would expect.
