# Is this strict inequality of integrals true?

Let $f,g$ be integrable functions on $[a,b] \subset \mathbb{R}$.

If it is known that $f(x) \leq g(x) \ \forall x \in [a,b]$, then $\int^{a}_{b} f(x) dx \leq \int^{a}_{b} g(x) dx$.

I was wondering about the following, as it would also seem to be true, but I'm having trouble proving it. Perhaps it's not true?

If $f(x) < g(x) \ \forall x \in [a,b]$, then is it the case that $\int^{a}_{b} f(x) dx < \int^{a}_{b} g(x) dx$.

• I think it is true. Commented Mar 24, 2016 at 6:37
• It's true if $f$ and $g$ are continuous. Commented Mar 24, 2016 at 6:39

Yes, I believe so. By considering $g-f$, it is equivalent to prove that if $f(x) >0$ then $$\int^b_a f(x) dx> 0.$$ To prove this, consider the sets $A_k = \{x \in [a,b] : f(x) > 1/k\}$, $k \in \mathbb N$. One of these sets must have positive (Lebesgue) measure since $[a,b] = \cup_k A_k$ has positive measure. Say $A_\ell$ has positive measure. Then $$\int^b_a f(x) dx\ge \int_{A_k} f(x) dx\ge \frac{\lambda(A_\ell)}{\ell} > 0,$$ where $\lambda$ is the Lebesgue measure on $\mathbb R$.