probability - Find the CDF with a given random variable

If $X$ is a random variable with a uniform distribution $U(0,1)$. Let $Y$ be a random variable defined with:

$Y=\begin{cases}X & ,0

Find the CDF for $Y$.

What bothers me is the 2nd case in which $Y=\frac{1}{2}$.
Should then $Fy$ be defined as $Fy=\int_{0}^{\frac{1}{2}}f(x)dx +\frac{1}{2} \int_{\frac{1}{2}}^{x}f(x)dx$

where $f(x)=1$ for this distribution?

Answer

It is easy to compute directly that
$$F_Y(y) = \Pr(Y\leq y)=\begin{cases}
0, & y\leq 0\\
y, & 0 < y <\tfrac12 \\
\tfrac13 + \tfrac23y, & \tfrac12 \leq y \leq 1\\
1, & y > 1
\end{cases}$$

Note that $F_Y(\tfrac12-) = \tfrac12$ but $F_Y(\tfrac12)=\tfrac23$. This is because $Y$ has a point mass at $Y=\tfrac12$ since $\Pr(Y=\tfrac12)=\tfrac16$. This means the graph of $F_Y(y)$ has a jump at $y=\tfrac12$ of size $\tfrac16$.