Let $D\subset\mathbb{R^2}$ a triangle which has the corners $(0,0),(1,0),(0,1)$ and $g: \mathbb{R} -> \mathbb{R}$ continuous. Then
$\int_Dg(x+y)dL^2(x,y)=\int_0^1tg(t)dt$
where $L^2$ is the Lebesguemeasure in $\mathbb{R^2}$.
So my idea is, to split $g=g_+-g_-$ then it is enough to show the equality for $\phi_I$ with $I=[t_1,t_2]\subset[0,1]$ and $\phi$ being the characteristic function of I since $g$ is continuous.
So let $F(x,y)=x+y$ then $\phi_I(F(x,y))=\phi_{F^{-1}(I)}$. Furthermore, let $D_t$ be a triangle with corners$(0,0) ,(t,0),(0,t). $ Then $\phi_{F^{-1}(I)}=D_{t_2}-D_{t_1}$
$\int_D\phi_I(F)dL^2=\int_{\mathbb{R}}\phi_D*\phi_{F^{-1}(I)}dL^2=\int_{\mathbb{R}}\phi_{F^{-1}(I)}dL^2=L^2(D_{t_2}-D_{t_1})=\frac{t_1^2}{2}-\frac{t_2^2}{2}=\int_{t_1}^{t_2}tdt=\int_{t_1}^{t_2}t*\phi_Idt$
Now letting $g_+=\sum\alpha _i\phi_{I_i}$ we get
$\int_Dg_+(F)dL^2=\sum\alpha _i\int_D\phi_{I_i}(F)dL^2=\sum\alpha _i\int_{I_i}t*\phi_{I_i}dt=\int_0^1tg_+(t)dt$
Doing this for $g_-$ then gives the equation.
This solution -if it is correct- is rather ugly in my opinion and I wonder if there's a better way to prove this.
A shorter way. By changing variables and letting $(t,s)=(x+y,x-y)$, we have that the new domain of integration is the triangle of vertices $(0,0)$, $(1,-1)$ and $(1,1)$, and $$\int_Dg(x+y)dL^2(x,y)=\int_{t=0}^1\int_{s=-t}^t g(t) \frac{dtds}{2}= \int_{t=0}^1g(t)\left(\frac{1}{2}\int_{s=-t}^t ds \right) dt=\int_{t=0}^1tg(t)\,dt,$$ where the Jacobian determinant is $|\partial(t,s)/\partial(x,y)|=2$.