Proving the Convolution of PDFs gives a PDF I am struggling with a question about the convolution of PDFs, in particular, proving that given two PDFs $f$ and $g$, then their convolution $f*g$, will also be a PDF. Proving non negativity is easy but I am struggling to show that the integrals has value $1$. Here is my proof so far:
Note that all these integrals are evaluated over the reals.
$$\int (f*g)(x)\text{d}x = \iint f(y)g(x-y) \text{d}y  \text{d}x$$
Let $ z = x-y $
\begin{align*}
\iint f(y)g(z)\text{d}y\text{d}z&= \int f(y) \text{d}y \int g(z) \text{d}z\\
&= 1
\end{align*}
I have the answer but I believe I have made an error with the changing of the limits of substitution when I introduce the new variable $z$. Any advice would be much appreciated.
 A: You are looking at a final result rather than where the convolution came from. Starting from an earlier point makes the proof easier.  
If $X$ and $Y$ are independent random variables with densities
$f$ and $g$ respectively, then
\begin{align}
P\{X+Y \leq z\} &= \int_{-\infty}^\infty P\{X+Y \leq z \mid Y = y\}g(y)\,\mathrm dy & \scriptstyle{\text{law of total probability}}\\
&= \int_{-\infty}^\infty P\{X \leq z-y \mid Y = y\}g(y)\,\mathrm dy\\
&=\int_{-\infty}^\infty P\{X \leq z-y\}g(y)\,\mathrm dy&
\scriptstyle{\text{independence of}~X~\text{and}~Y}\\
P\{X+Y \leq z\} &=\int_{-\infty}^\infty 
\left[ \int_{-\infty}^{z-y} f(x)\,\mathrm dx\right]g(y)\,\mathrm dy
\tag{1}
\end{align}
Note that the left side of $(1)$ has limiting value $1$ as
$z \to \infty$; in fact, the value of that double integral
on the right is the CDF of the random variable $Z = X+Y$.
The Fundamental Theorem of Calculus applied to $(1)$
gives us
$$f_{X+Y}(z) = \frac{\mathrm d}{\mathrm dz}P\{X+Y \leq z\}
= \int_{-\infty}^\infty f(z-y)g(y)\,\mathrm dy
= f\star g$$
and so
$\displaystyle \int_{\mathbb R} f\star g = 1$ as you wish to prove.
