A question about lower Riemann integral

Question

Body: I've been wondering if $\left|\int_{\overline{a}}^{b} f(x) \: dx\right| \leq \int_{\overline{a}}^{b} |f(x)| \: dx$, where all the integrals are Lower Riemann Integrals. I know, by integral properties, that $\left|\int_{a}^{b} f(x) \: dx\right| \leq \int_{a}^{b} |f(x)| \: dx$. I've also proved that if $f: [a,b] \to \mathbb{R}$ is a bounded function, $\left|\int_{a}^{\overline{b}} f(x) \: dx\right| \leq \int_{a}^{\overline{b}} |f(x)| \: dx$, where all the integrals are Upper Riemann Integrals. So my question is if does it follow for lower Riemann integrals. Any hints would be highly appreciated!

Answer 1

It is true that

$$\tag{1}\underline{\int}_a^bf(x) \, dx \leqslant \underline{\int}_a^b |f(x)| \, dx$$

holds for any bounded function. However, it is not true that for any bounded function we have

$$\tag{2}\left|\underline{\int}_a^bf(x) \, dx\right| \leqslant \underline{\int}_a^b |f(x)| \, dx$$

To prove (1), note that, on any partition interval $I$ we have $\inf_{x \in I}f(x) \leqslant \inf_{x \in I} |f(x)|$ and the lower Darboux sums satisfy $L(P,f) \leqslant L(P,|f|$).

Thus, for any partition $P$,

$$L(P,f) \leqslant L(P,|f|) \leqslant \underline{\int}_a^b |f(x)| \, dx,$$ and

$$\sup_PL(P,f) =\underline{\int}_a^b f(x) \, dx \leqslant \underline{\int}_a^b |f(x)| \, dx$$

A counterexample for (2) is the Dirichlet-type function

$$f(x) = \begin{cases}-1, &x \in \mathbb{Q}\cap [0,1]\\0, & x \in [0,1]\setminus\mathbb{Q}\end{cases},$$

where $\left|\underline{\int}_0^1 f(x) \, dx\right| = |-1| = 1$ and $\underline{\int}_0^1 |f(x)| \, dx= 0$.

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It is somewhat surprising that the triangle inequality (2) does not always hold for lower integrals while it holds for upper integrals (as well as for the Riemann integral).

Why can't we prove the inequality for lower integrals by replacing $f$ with $-f$ and applying the inequality for upper integrals as suggested in a comment?

Since $\inf_{x \in I} f(x) = - \sup_{x\in I} [-f(x)]$, it follows that $L(P,f) = -U(P,-f)$, and, hence,

$$\underline{\int}_a^bf(x) \, dx=\sup_P L(P,f) = \sup_P [- U(P,-f)] = - \inf_P U(P,-f) = - \overline{\int_a}^b [-f(x)] \, dx$$

Taking the absolute value, and applying the triangle inequality for upper integrals we get

$$\left|\underline{\int}_a^bf(x) \, dx\right|= \left|- \overline{\int_a}^b [-f(x)] \, dx\right|= \left| \overline{\int_a}^b [-f(x)] \, dx\right| \leqslant \overline{\int_a}^b |-f(x)| \, dx = \overline{\int_a}^b |f(x)| \, dx,$$

which does not prove (2).