# Disjoint support of functions leads to simple integral

Suppose $f: \mathbb{R^3} \rightarrow \mathbb{R}_+$ and $g: \mathbb{R} \rightarrow \mathbb{R}_+$ with $supp(f)\nobreakspace \cap supp(g^{3}) \approx 0$. Is there a simple argument why then

$\int \nobreakspace \mid f(x_1,x_2,x_3) - \prod_{i=1}^3 g(x_i)\nobreakspace \mid dx_1 dx_2 dx_3 = 2$

$\mid \cdot \mid$ denotes the euclidean norm.

• What are the assumptions on $f$ and $g$. Are they density functions in 3 dimensions and 1 dimension? – Kabo Murphy May 17 '18 at 9:26
• Yes exactly, they are densities – Jack_Stiller10 May 17 '18 at 9:57

If $f$ and $g$ have disjoint supports then $\int |f-g| =\int |f| + \int|g|$ as seen by integrating over the supports $A$ and $B$ of the two functions and then over $\mathbb R^{3} \setminus (A \cup B)$. Hence the given integral is 1+1=2.