How to find limits of integration on a convolution of CRVs In finding the convolution of two independent and continuous random variables, I am struggling with limits of integration.  I cannot seem to figure out over what intervals the probability density function $f_{Z}(a) = f_{X+Y}(a)$ breaks out to.
The most basic example is where $f_{X}(a)$ and $f_{Y}(a)$ are both uniform over $[0,1]$ and independent.  The intervals for $f_{Z}(a)$ are $(0,1)$ and $(1,2)$.  But why? There are also more complicated cases such as two exponential  R.V.s (say with parameters $\lambda$ and $2\lambda$) or an exponential and a uniform, (say $\lambda$ and $[0, 1]$), etc.
For reference:
$$f_{x+y}(a) = \int_{-\infty}^{\infty} f_X(a-y)f_Y(y)~dy$$
Once I can set it up, the integration is (usually) no trouble.
 A: One easy way to think about this is to look at $f_X$ and $f_Y$ and figure out where exactly these pdfs vanish, thus not contributing anything to the convolution.
Let's look at the uniform case. Since the uniform is defined on $[0,1]$, obviously anything outside of this won't make sense. So now we have $$f_{X+Y}(a)=\int_{-\infty}^\infty f_X(a-x)f_Y(x) \mathrm{d}x=\int_{0}^1 f_X(a-x)f_Y(x) \mathrm{d}x$$
Now, $f_Y$ is "safe", since it is defined on $[0,1]$. We need to ensure the same thing for $f_X$, that is, we want $0 \leq a-x \leq 1$. From this, one gets $x \leq a$ and $x \geq a-1$. This is what breaks the integral into two pieces. The first piece tells you that you should have $$f_{X+Y}(a)=\int_0^a \mathrm{d}x = a$$ as long as $x \leq a$, but $x \in [0,1]$ according to the convolution domain, implying $a \in [0,1]$ as well.
From the second inequality we get $$f_{X+Y} = \int_{a-1}^1 \mathrm{d}x = 2-a$$ as long as $x \geq a-1$. Since $x \in [0,1]$, then $a \in [1,2]$. This reconciles with the obvious observation that the sum of two r.v. in $[0,1]$ must live in $[0,2]$.
There's your triangular distribution: $a$, when $a \in [0,1]$ and $2-a$ when $a \in [1,2]$.
If you want to convolve, say, two exponentials, then your convolution domain reduces to $[0, \infty)$. Your inequality should say that $0 \leq a-x < \infty$. Can you take it from here?  
