Finding the distribution of the rounding error random variable Suppose that a non-negative random variable $X$ has a distribution function $F(x)$, and that $Y$ is the rounding error if $X$ is rounded off to the nearest integer below. Show that $Y$ has the distribution function
$$\sum_{j=0}^\infty [F(j+y) - F(j)] $$ where $$ (0 \le y < 1)$$
My thoughts:
$Y = X- \lfloor X\rfloor$
$P(Y \le y) = P(X-\lfloor X\rfloor \le y)$
So I need to find the density function of $X-\lfloor X\rfloor$.
I first look the density function of $\lfloor X\rfloor$.
$P(\lfloor X\rfloor \le k) = P(0 \le X < k+1)$ where $k$ is an non-zero integer.
$P(0 \le X < k+1) = \int^{k+1}_0f(x)dx  = $. Then I don't how should I continue;
 A: $$\begin{align}
P(Y \le y) &= P(X-\lfloor X\rfloor \le y) \\
&= P\left(~~  X - 0 \le y \quad\text{or}\quad X - 1 \le y \quad\text{or}\quad X - 2 \le y ~\ldots ~~\right) \\
&= \left[ F(0+y) - F(0)\right] + \left[ F(1+y) - F(1)\right] + \left[ F(2+y) - F(2)\right]+ \ldots
\end{align}$$
the probability can be split like that and added since $X$ is non-negative and the cases are mutually exclusive.
Personally I take the above as just a reasonable and intuitive way to "look at" and "understand" the desired result but not a rigorous derivation. This $\sum_{j=0}^\infty [F(j+y) - F(j)]$ is easy to visualize: just cut the density of $X$ at the integers and shuffle/align the curve segments. Note that it indeed starts at $P = 0$ at $y=0$ and goes up to $P \to 1$ as $y \to 1^{-}$.
I've been waiting for someone to give a formal proof, e.g. following the OP's line of thought treating it as $Y = X + W$ where $W \equiv -\lfloor X\rfloor$ and doing the convolution or MGF after finding out the density of $W$ (which is essentially a discrete pmf). This route seems technically tedious and might NOT be illuminating at all, but I believe it is possible and can be useful for more complicated situations.
Anyway, here's my two cents and I hope this question stays visible for longer and someone else more capable can contribute. 
