Similar sets but different integrals I have a question concerning These two similar sets:
\begin{align*}
A&=\{(s,y)\in\mathbb{R}_+ \times \mathbb{R}_+:s+y\leq t\}\\
B&=\{(w,z)\in\mathbb{R} \times \mathbb{R}:w+z\leq t\}\\
\end{align*}
I found two integral over each of these sets. 
$$I(B) = \int_{-\infty}^\infty \int_{-\infty}^{t-z} f(w,z) \mathrm{d}w \mathrm{d}z$$
and
$$I(A)=\int_{0}^t \int_{0}^{t-s} f(s,y) \mathrm{d}y \mathrm{d}s$$
But why is the outer integral different? Why is it not
$$I(A)=\int_{0}^\infty \int_{0}^{t-s} f(s,y) \mathrm{d}y \mathrm{d}s$$
I don't understand this. Or is one wrong?
 A: There is a significant difference. $A$ is bounded while $B$ is unbounded.
In the set $B$ you have just one restriction for $w$ and $z$, namely $w+z\leq t$ Therefore you can choose one variable free in whole $\mathbb{R}$ while the other is bounded from above by the restriction. This yields
$$
I(B)=\int_{-\infty}^\infty\int_{-\infty}^{t-z}f(w,z)dwdz.
$$
But the set $A$ is just a small triangle. You can see it as three restriction:
$$
A=\{(s,y)\in\mathbb{R}\times\mathbb{R}~:~0\leq s,~0\leq y,~s+y\leq t\}.
$$
You see that $s$ as well as $y$ are bounded. Starting with $s$ you get $$
0\leq s\underbrace{\leq}_{\text{ since }y\geq 0}s+y\leq t. 
$$
And therefore the restriction for $y$ has to be $$0\leq y\leq t-s$$ and we get
$$
I(A)=\int_0^t\int_0^{t-s}f(s,y)dyds.
$$
A: With loss of generality, assume $t\geq 0$. Region $A$ is given graphically as 
 
while region $B$ is given as 
.  
This means 
\begin{align*}
I(A) &= \int_{0}^{t} \int_{0}^{-s+t} f(s,y)\: dyds \\
&= \int_0^t \int_0^{-y+t} f(s,y)\: dsdy \\  
\end{align*}
and 
\begin{align*}
I(B) &= \int_{-\infty}^{\infty} \int_{-\infty}^{-z+t} f(w,z)\: dwdz \\
&= \int_{-\infty}^{\infty} \int_{-\infty}^{-w+t} f(w,z) \: dzdw. \\  
\end{align*}
A: Set A does not include points greater than t along the s and y axes
