Convolution of continuous and discrete distributions Assume we have two random variables $X$ and $Y$, such that $X \sim P(x)$ and $Y \sim G(y)$.
We ask, what is the distribution of $Z = X+Y$.
If both of the distributions of $X$ and $Y$ are discrete, the distribution of $Z$ is given by the convolution of the two, ie. $$F(z) = \sum_{x \in \mathbb Z} P(x) G(z-x).$$
Similarly, if the distributions are continuous, the convolution is $$F(z) = \int_{-\infty}^{\infty} P(x) G(z-x) dx.$$
Now, what if other, say $P(x)$ is discrete and the other continuous?
Take as an example probably the simplest one, $X \sim \text{Bernoulli}(1/2;x)$ and $Y \sim U(0,1)$.
If one thinks of this, $Z$ has probability $1/2$ of being between zero and 1 and same probability of being between one and two. Thus one could say immediately that $Z \sim U(0,2).$ 
The way I have seen this done elsewhere involves the use of a distribution called Dirac delta function $\delta(x)$, which is zero everywhere else except at zero but the integral over all reals is $1$.
Using this, we can make $\text{Bernoulli}(1/2;x)$ a continuous distribution, namely $$\text{DeltaBernoulli(1/2;x)} \sim \text{Bernoulli}(1/2;x)\delta(x)+\text{Bernoulli}(1/2;x)\delta(x-1).$$ (Subquestion: is this convolution of the two distributions or something else?)
Using this, lets perform the convolution $$M(z) = \int_{-\infty}^{\infty} 1*\text{PDF}(\text{DeltaBernoulli}(1/2;z-x)) dx = \begin{cases}
   1/2 & 0\le z< 1 \vee 1 < z \le 2 \\
   1       &z = 1 \\
   0   &\text{elsewhere}
  \end{cases}
 $$
This is almost $U(0,2)$ and in any case has no real differences to it.
This method seems a tad clumsy, especially if one would want to convolute, say, normal distribution and Poisson distribution.
Are there other ways to do this?
 A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{{\rm F}\pars{z} = \int_{-\infty}^{\infty}{\rm P}\pars{x}{\rm G}\pars{z-x}
     \,\dd x}$

If ${\rm P}\pars{x}$ is discrete, we can write it as
$\ds{{\rm P}\pars{x} = \sum_{n}P_{n}\,\delta\pars{x - x_{n}}}$ where $\ds{\braces{P_{n}}}$ is the probability of $x_{n}$. Then, 
$$\color{#0000ff}{\large%
{\rm F}\pars{z} = \int_{-\infty}^{\infty}{\rm P}\pars{x}{\rm G}\pars{z-x}\,\dd x
=
\sum_{n}P_{n}\,{\rm G}\pars{z - x_{n}}}
$$

A: If the random variables are independent, you can use the characteristic functions of the Random variables since if:
$$Sn=\sum_{i=1}^{n}a_iX_i$$
then
$$\phi_{S_n}(t)=\prod_{i=1}^n\phi_{X_i}(a_it)$$
So for the case given
$$\phi_{X}=1-\frac{1}{2}+\frac{1}{2}e^{it} \text{ and } \phi_Y=\frac{e^{it1}-e^{it0}}{it(1-0)}$$
so
$$\phi_Z=\frac{e^{2it}-e^{it0}}{it(2-0)}$$
which is $U(0,2)$, not just close to.
