Show that, $\int_A fg\ge \frac{1}{|A|}\int_A f\int_A g$ 
If $A\subset \mathbb{R}$ and $f,g :A\to\mathbb{R}$ are monotone increasing, Lebesgue integrable functions, then $$\int_A f(x)g(x)dx\ge \frac{1}{|A|}\int_A f(x)dx\int_A g(x)dx.$$

It looks like the Jensen-Inequality, but i think we can't apply it in this case. I don't know how to define such a convex function.
But since $f,g$ are integrable, using Fubini we get on the RHS:
$\int f(x)dx\int g(x)dx=\int\int f(x)g(y)dxdy$
this is a double integral, and on the LHS we have only one variable, can we do
$\int f(x)g(x)dx=\dfrac{1}{2}\int\int f(x)g(x)+f(y)g(y)dxdy$
or is my approach completely wrong ?
Thanks in advance.
 A: I tried to use mean value theorem but from the hint I got it.
The function $h(x,y) = (g(y)-g(x))(f(y)-f(x)) \geq 0$, since (by the monotone increasing of $f$ and $g$) if $y>x$, both terms in the product are non-negative and if $y\leq x$, both terms are non-positive. The product is non-negative in both cases.
(Let's use the interval [0,1] as a domain for $f$ and $g$ in the claim, maybe it can also be formulated by taking average integral.)
So if we integrate $h$ over $[0,1]^2$ we get something non-negative:
$$0\leq \int_0^1\int_0^1 h(x,y)dxdy
= \int_0^1\int_0^1 {g(y)f(y) -g(y)f(x)-g(x)f(y)+g(x)f(x)}dxdy $$
$$= 1\cdot\int_0^1 g(y)f(y)dy -\int_0^1\int_0^1 g(y)f(x)dxdy-\int_0^1\int_0^1 g(x)f(y)dxdy+ 1\cdot\int_0^1 g(x)f(x)dx $$
$$= 2\int_0^1 f(x)g(x)dx - 2\int_0^1f(x)dx \int_0^1g(x)dx$$
Hence the claim follows. You were on the right track with Fubini. The trick is to notice that $h\geq0$.
EDIT: fixed typos and did some extra steps.
For a domain $A$ with finite measure the claim should be:
$$\int_A f(x)g(x)dx \geq \frac{1}{|A|}\int_A f(x)dx\int_A g(x)dx$$
This can be seen as before with $A^2$ as the domain of $h$. For sets with infinite measure, I guess you can just do approximation argument.
A: You have not specified the domain of integration. If you take the domain to be $[0,\, 2]$ and $f(x) = g(x) = x$ you violate the inequality in question.
