Independence between a constant random variable and another random variable. Intuitively, I understand that if $Y$ is a constant random variable and $X$ is another random variable, then $X$ and $Y$ are independent. 

However, I can't make a formal proof because I can't show that their joint density function are the product of two functions that rely only on x's and y's respectively or using similar methods. 

(What is the density function of a constant random variable for example?)

Can you give me a hint in order to make a proof?
$X$ and $Y$ are independent if and only if $P(X\in A, Y\in B)=P(X\in A)P(Y\in B)$ for all $A,B$.

Assume $Y=y$ for some $y\in\mathbb{R}$. Then $$P(X\in A, Y\in B)=\left\{\begin{array}{ll}0&amp;\mathrm{if}\,\,y\notin B,\\ P(X\in A)&amp;\mathrm{if}\,\,y\in B\end{array}\right.$$

But notice $P(Y\in B)=1$ if $y\in B$ and $P(Y\in B)=0$ if $y\notin B$.
Show a more general result, that if $Y$ is a constant random variable with value $c$ with probability 1, it is independent to any random variable $X$. Essentially, the result we need is from random statement 24. Let $k \neq c$ be a constant.
$\Pr(Y=c)=1$, so $\Pr (B, Y=c) = \Pr (B)$ for all events $B$. In particular, setting $B=\{ X=x \}$ gives $ \Pr (X=x) \Pr (Y=c) = \Pr (X=x) = \Pr(X = x, Y = c)$.
$\Pr(Y=k)=0$, so $\Pr (B \cup (Y=k)) = \Pr (B)$ for all events $B$. In particular, setting $B=\{ X=x \}$ and using the inclusion-exclusion rule gives $ \Pr(X = x, Y = k) = \Pr (X=x) + \Pr (Y=k) - \Pr (X=x \cup Y=k) = \Pr (X=x) + 0 - \Pr (X=x) = 0 = \Pr (X=x) \Pr(Y=k)$.