How to prove Fubini-Tonelli for "Cavallieri" like sets?

Question

One can sort of combine Cavalieri's principle and Fubini's theorem in a way that the following statement holds: $$ \int_T f(x,y) d(x,y) = \int_{\mathbb{R}} \int_{T_x} f(x,y) dx = \int_0^1 \int_0^{1-y} f(x,y) dx dy $$ where $$ T = conv\{(0,0),(0,1),(1,0)\}, T_x=\{y \in \mathbb{R} : (x,y) \in T\} $$ and $f$ continuous. I am struggling to find a rigorous proof for this. Can you help me? If $f=1$, then we have Cavalieri's principle. And if we have a set $T=T_1 \times T_2$, then a similar statement gives us Fubini's theorem (the generalized version). But I struggle to connect those two.

Answer 1

I've figured it out: $$ \int_T f(x,y)d(x,y) = \int_{\mathbb{R}^2} f(x,y) \mathbb{1}_T(x,y) d(x,y) =\int_\mathbb{R} \int_\mathbb{R} f(x,y) \mathbb{1}_T(x,y) dx dy = \\\int_\mathbb{R} \int_\mathbb{R} f(x,y) \mathbb{1}_{[0,1]}(y) \mathbb{1}_{[0,1-y]}(x) dx dy = \int_\mathbb{R} \mathbb{1}_{[0,1]}(y) \int_\mathbb{R} f(x,y) \mathbb{1}_{[0,1-y]}(x) dx dy = \\\int_{[0,1]} \int_{[0,1-y]} f(x,y) dx dy $$