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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 $\ge$.

BlueNight
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  • Also: https://math.stackexchange.com/q/553031/42969, https://math.stackexchange.com/q/360264/42969, https://math.stackexchange.com/q/357472/42969 – Martin R Mar 07 '19 at 14:00

2 Answers2

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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 $$\int_{[a,b]} f \ge \int_{[x_0 - \delta, x_0 + \delta]} f \ge 2\delta \cdot \frac{f(x_0)}2 = \delta f(x_0) > 0$$

mechanodroid
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If the function in continuous, then you can use a lower-bounding argument as follows:

  1. If $f(x) > 0$, there is a small enough $\epsilon>0$ such that $f(x)>\epsilon$ for all $x\in [a,b]$,
  2. You can lower bound $$\int_a^b f(x) \mathrm d x \geq \int_a^b \epsilon \mathrm d x = \epsilon (b-a),$$
  3. 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 $$\int_a^b f(x) \mathrm d x \geq \int_{a'}^{b'} f(x) \mathrm d x > 0.$$

Riccardo Sven Risuleo
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    Note that the function is not assumed to be continuous. – Martin R Mar 07 '19 at 13:52
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    How about $$f(x) = \begin{cases} x-a, &\text{ if } x \in \langle a,b] \ 1, &\text{ if } x= a\end{cases}$$ Then $f$ is Riemann-integrable since it is continuous on $\langle a,b]$, and $\lim_{x \to a^+} f(x) = 0$. – mechanodroid Mar 07 '19 at 14:01
  • Ah true! I implicitly assumed that it was continuous. – Riccardo Sven Risuleo Mar 07 '19 at 14:19
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    $f$ can be Riemann-integrable without being continuous on any interval of positive length. – Martin R Mar 07 '19 at 14:29
  • Yep, for example a variant of Thomae's function: $$f(x) = \begin{cases} 1+\frac1q, &\text{ if } x \in \mathbb{Q}, x = \frac{p}q \text{ in lowest terms}\ 1, &\text{ if } x \notin \mathbb{Q}\end{cases}$$ Its set of discontinuities is $[a,b] \cap \mathbb{Q}$ so it is Riemann-integrable but not continuous on any inverval in $[a,b]$. – mechanodroid Mar 07 '19 at 19:45