Double integral of a single variable function $$\int_{\mathbb{R}}  \left(\int_{\mathbb{R}} f(x)dx \right) g(x) dx$$
Can this integral be reduced to 
$$  \int_{\mathbb{R}} f(x)g(x)dx $$
 A: No, in general
$$\int_{\mathbb{R}}  \Big(\int_{\mathbb{R}} f(x)dx \Big) g(x) dx\not=\int_{\mathbb{R}} f(x)g(x)dx.$$
Take for example $f(x)=g(x)=x$ for $x\in [-1,1]$ and zero elsewhere. Then, by symmetry, the LHS is zero whereas the RHS is not zero.
However, if the integrals are convergent then, by linearity,
$$\int_{\mathbb{R}}  \Big(\int_{\mathbb{R}} f(x)dx \Big) g(x) dx=\Big(\int_{\mathbb{R}} f(x)dx \Big)\cdot\Big(\int_{\mathbb{R}}  g(x) dx\Big).$$
A: No it cannot be reduced. $\Big(\int_{\mathbb{R}} f(x)dx \Big)$ is a constant, let's say $k$, which reduces the integral to $k\int_{\mathbb{R}} g(x)dx $
However, this proves the following result:
$$\int_{\mathbb{R}}  \left(\int_{\mathbb{R}} f(x)dx \right) g(x) dx=\left(\int_{\mathbb{R}} f(x)dx \right)\left(\int_{\mathbb{R}} g(x)dx \right)$$
A: The expression $$\int_{\mathbb R} \left(\int_{\mathbb R} f(x)dx\right)g(x)dx$$
is a poorly written expression. It is not techincally wrong, but it is confusing. Following the strict definition of an integral, technically speaking, the expression equivalent to 
$$\int_{\mathbb R} \left(\int_{\mathbb R} f(y)dy\right)g(x)dx$$
which converges if $$\int_{\mathbb R}g(x)dx$$ converges. This is because 
$$\int_{\mathbb R} \left(\int_{\mathbb R} f(y)dy\right)g(x)dx=\left(\int_{\mathbb R} f(y)dy\right)\left(\int_{\mathbb R}g(x)dx\right)$$
It is not equal to $\int_{\mathbb R} f(x)g(x)dx$.
