Does this inequality for sums hold for integrals too?

Question

Let $f,g:I \to \mathbb R$ be positive functions on a finite set, then $$ \left(\sum_i f(i)\right)\left(\sum_i g(i)\right) \geq \sum_i f(i)g(i) $$ The proof is rather trivial: by multiplying out the sums, you can write the LHS as $\sum_i f(i)g(i) +$ crossterms, which are positive by assumption. But when $f,g$ are (still positive) functions $X \to \mathbb R$ from a measure space $X$ then I can't find a reference for it: $$ \left(\int_x f(x)dx\right)\left(\int_x g(x)dx\right) \geq \int_x f(x)g(x)dx $$ This looks like an Holder-type inequality $\|f\|_1\|g\|_1 \geq \|fg\|_1$. Can I use Fubini to conclude $\left(\int_x f(x)dx\right)\left(\int_x g(x)dx\right) = \int_{x,y}f(x)g(y)dxdy$ and then reason like above? Or can you suggest a different argument?

Answer 1

It is false. If $f=g=1_E$ the inequality says $\lambda (E)^{2} \ge \lambda (E)$ (where $\lambda$ is Lebesgue measure). This is true only if $\lambda (E) \ge 1$.