Does replacing $\ge$ by $>$ lead to a strictly inequality too? (integral inequality) I know that if $f:[a,b] \to \mathbb{R}$ is an integrable function such that $f(x) \ge 0,~\forall~x \in [a,b],$ then $$\int\limits_a^b f(x) \mathrm{d}x \ge 0.$$
What happens if we replace the condition $f(x) \ge 0$ by the condition $f(x)>0$? Will the last inequality be also strict?
I tried to use Riemann sums but taking the limit turns $>$ into $\ge$.
 A: The answer is yes.
Since $f$ is Riemann-integrable on $[a,b]$, the Lebesgue integrability criterion implies that the set of discontinuities $D$ of $f$ has Lebesgue measure zero.
In particular, there exists $x_0 \in \langle a,b\rangle$ such that $f$ is continuous at $x_0$.
Using $\varepsilon = \frac{f(x_0)}2 > 0$ in the definition of continuity yields that there exists $\delta > 0$ such that for all $x \in [a,b]$ holds $$x \in [x_0 - \delta, x_0 + \delta] \implies f(x) > \frac{f(x_0)}2$$
In particular
$$\int_{[a,b]} f \ge \int_{[x_0 - \delta, x_0 + \delta]} f \ge 2\delta \cdot \frac{f(x_0)}2 = \delta f(x_0) > 0$$
A: If the function in continuous, then you can use a lower-bounding argument as follows:


*

*If $f(x) > 0$, there is a small enough $\epsilon>0$ such that $f(x)>\epsilon$ for all $x\in [a,b]$,

*You can lower bound
$$\int_a^b f(x) \mathrm d x \geq \int_a^b \epsilon \mathrm d x = \epsilon (b-a),$$

*If $b\neq a$, $b-a>0$, and you have the result.



EDIT:
If it is not continuous, take an interval $[a',b']$ inside $[a,b]$ such that $f(x)$ is continuous on $[a',b']$, then you can use the same argument to say that
$$\int _{a'}^{b'} f(x) \mathrm d x >0,$$
then use the nonnegativity of $f(x)$ to say that
$$\int_a^b f(x) \mathrm d x \geq \int_{a'}^{b'} f(x) \mathrm d x > 0.$$
