Comparison between integrals

Question

Let's say - $F(x) > G(x)$ for all $x$ in $E$.

So is it true that ? $$\int F(x)\,dx > \int G(x)\,dx$$ or I can use this statement only when the two functions are non-negative functions.

Thanks!

Answer 1

The statement in your question is a consequence of the statement for nonnegative functions and linearity of the integral.* If $F,G$ are integrable on $E$ and $F(x)>G(x)$ for all $x \in E$, then $F(x)-G(x)>0$ for all $x \in E$. Now $F(x)-G(x)$ and $0$ are both non-negative integrable functions on $E$, and $$\int_E (F(x)-G(x))\ dx \geq \int_E 0 \ dx = 0.$$ By linearity, $\int_E (F(x)-G(x)) \ dx = \int_E F(x) \ dx - \int_E G(x) \ dx$, so the above implies $$\int_E F(x) \ dx \geq \int_E G(x) \ dx.$$

*As noted in the comments, the inequality is not strict if $E$ has measure zero. If $E$ has positive measure, let $E_n=\{x \in E : F(x)-G(x)>1/n\}$, and note that $E=\bigcup_{n=1}^\infty E_n$. Hence some $E_n$ must have positive measure, and we have \begin{align*} \int_E(F(x)-G(x))\ dx &\geq \int_{E_n}(F(x)-G(x)) \ dx\\ &\geq \int_{E_n} \frac{1}{n} \ dx \\ &=\frac{1}{n}\cdot m(E_n)\\ &>0. \end{align*} So in the case that $E$ has positive measure, the inequality is strict.