Let $X$ a real random variable with density $f_X$ and $Y=g(X)$ with $g$ measurable.
I have some trouble understanding the following statement :
" If $g(x)=a$ with $a$ a constant, for all $x$, the distribution of $Y$ is the Dirac mass in $a$, which doesn't have a density "
My questions :
1 - How can we translate this statement into formulas?
2 - Why doesn't the density exist?
On
Unpack the variable $Y$: If $c < d$ are real numbers, then
$$
P(c < Y < d) = P(c < g(X) < d) = P(c < a < d)
$$
This is 1 if $a$ is between $c$ and $d$, and $0$ if it isn't.
So by extension, the distribution of $Y$ is identical to the distribution of a point mass centered at $a$:
$$
\mu_Y(E) = P(Y \subseteq E) = \begin{cases} 1 & a \in E \\ 0 & a \notin E \end{cases}
$$
The distribution of $Y$ doesn't have a density function. To see why, suppose it did have a density $f_Y$. let $E$ be the set of numbers $\neq a$. Then for any measurable set $E'\subseteq E$, $$ \int_{E'} f_Y\,dx = \mu_Y(E') = 0 $$ since $a \notin E'$. This means that $f_Y = 0$ almost everywhere in $E$. Replacing $a$ (a set of measure zero by itself), we still have $f_Y = 0$ almost everywhere in $\mathbb{R}$. But $$ \int_{\mathbb{R}} f_Y\,dx = \mu_Y(\mathbb{R}) = 1, $$ a contradiction.
Let $(\Omega,\mathcal A,\Pr)$ denote the probability space that serves as domain for random variable $X:\Omega\to \mathbb R$. Then $\{Y\in A\}=\{\omega\in\Omega\mid g(X(\omega))\in A\}=\Omega$ if $a\in A$ and $\{Y\in A\}=\varnothing$ otherwise.
1) Let $P_Y$ denote the probability meausure on $(\mathbb R,\mathcal B)$ prescribed by $A\mapsto\Pr(Y\in A)$.
That leads to:$$P_Y=\delta_a$$
This because $P_Y(A)=\Pr(Y\in A)=1$ if $a\in A$ and $P_Y(A)=\Pr(Y\in A)=0$ otherwise.
2) The distribution (it exists all right) will have function $F(x)=1_{[a,\infty)}(x)$ as cumulative distribution function. In order to have a density function the CDF must at least be continuous, which is not the case.