Real Analysis Riemann Integration - Strict Monotonicity for Integrals

If $f,g$ are Riemann integrable on $[a,b]$, and $f(x) < g(x)$ for all $x \in [a,b]$, prove that $$ \int_a^b f(x) \,dx < \int_a^b g(x) \,dx$$

This is a strict inequality. I know how to prove the monotonicity of integrals with the non strict inequality, but I am not sure how to do this.

Solution 1:

Let $h(x) =g(x) - f(x) $ and then $h$ is Riemann integrable on $[a, b] $ and hence we know that $h$ must be continuous somewhere in interval $[a, b] $. Let $h$ be continuous at some $c\in[a, b] $ and since $h(c) >0$ there is a subinterval $[p, q] $ such that $c\in[p, q] \subseteq[a, b] $ and $h(x) >h(c) /2$ for all $x\in [p, q] $. And since $h$ is positive on $[a, b] $ we have $$\int_{a} ^{b} h(x) \, dx\geq \int_{p} ^{q} h(x) \, dx\geq (q-p) \cdot\frac{h(c)} {2}>0$$ and thus $\int_{a} ^{b} f(x) \, dx<\int_{a} ^{b} g(x) \, dx$.

Solution 2:

It suffices to show $f(x) >0$ for all $x$ $\in [a,b] \implies \int_a^b f(x) \,dx>0$.

Since you already know $\int_a^b f(x) \,dx\geq 0$, we suppose $\int_a^b f(x) \,dx=0$. The integral implies except a measure zero set, $f=0$ on $[a,b]$. since $[a,b]$ has positive measure, $\exists x_0\in [a,b]$ s.t. $f(x_0)=0$, we get a contradiction.