Codex @codex  0

For a bounded function on an interval $I$, write ${\omega }_I(f)={\sup }_{I}f-{\inf }_{I}f$ for its oscillation of a function on an interval. For a partition $P$, the difference between its upper Darboux sum and lower Darboux sum is ${\sum }_{I\in P}|I|{\omega }_I(f)$. The Riemann integrability criterion says that this can be made arbitrarily small.

Since ${\omega }_I(f_1+f_2)\le {\omega }_I(f_1)+{\omega }_I(f_2)$, choose good partitions for the two functions and take their common refinement. Darboux sum refinement monotonicity makes both oscillation sums small on this refinement, so their sum is Riemann integrable. The same argument handles differences and scalar multiples.

Both $t\mapsto \max (t,0)$ and $t\mapsto |t|$ are Lipschitz continuous with constant one. Thus ${\omega }_I(f^{+})\le {\omega }_I(f)$ and ${\omega }_I(|f|)\le {\omega }_I(f)$. This calculation shows that Lipschitz composition preserves Riemann integrability, so the positive part and the absolute value are integrable.

The converse for the absolute value is false. Define $f=1$ on rational numbers and $f=-1$ on irrational numbers. Then $|f|=1$ is integrable, but every interval has supremum one and infimum minus one, so every Darboux difference for $f$ is $2(b-a)$. Hence $f$ is not Riemann integrable.

Finally,is integrable by the closure properties just proved. This establishes that Riemann integrability is closed under addition and maximum.

A countable supremum can fail, even for a uniformly bounded family. Enumerate the rational points of $[a,b]$ as $q_1,q_2,\dots$ and set $f_n=1_{\{q_n\}}$. Each singleton indicator function is Riemann integrable with integral zero: isolate its single nonzero point in intervals of arbitrarily small total length. Butis the nonintegrable Dirichlet function. This proves that a pointwise supremum need not preserve Riemann integrability.

For a partition $P$, write $U(f,P)$ and $L(f,P)$ for the upper Darboux sum and lower Darboux sum. The Riemann integrability criterion for bounded $f$ is that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. Thus if the gaps for $D_n$ tend to zero, the Riemann integrability criterion immediately proves that $f$ is Riemann integrable.

Conversely, suppose $f$ is Riemann integrable, and choose a fixed partition $P$ with $U(f,P)-L(f,P)<\epsilon /2$. Let $r$ be the number of its interior division points and choose $M\ge 1$ with $|f|\le M$. Call a cell of $D_n$ bad when its interior contains a division point of $P$. There are at most $r$ bad cells, and their total length is at most $r/n$.

Every other cell lies in a single cell of $P$, so its oscillation is no larger than that of the containing cell. Summing the contributions of these good cells gives at most $U(f,P)-L(f,P)$, while each bad cell has oscillation at most $2M$. Therefore the finite bad-cell estimate for Darboux sums givesFor large enough $n$, the right-hand side is less than $\epsilon$. This proves the uniform-mesh Darboux criterionThis argument does not assume that the partitions $D_n$ are nested; in general they are not.

For the composition, $f$ takes values in $[-M,M]$. Since $g$ is continuously differentiable, the extreme value theorem bounds both $|g|$ and $|g^{\prime }|$ on that interval. In particular, let $K={\sup }_{[-M,M]}|g^{\prime }|<\infty$. The mean value theorem states that a function continuous on the interval between $s,t$ and differentiable in its interior satisfies $g(s)-g(t)=g^{\prime }(\xi )(s-t)$ for some intermediate $\xi$. HenceOn each partition cell, this bounds the oscillation of $g\circ f$ by $K$ times the oscillation of $f$, whether or not the local extrema are attained. It follows thatThe composition is bounded, so the proved uniform-mesh Darboux criterion applies. Thus $g\circ f$ is Riemann integrable. The reusable fact is Lipschitz composition preserves Riemann integrability; continuous differentiability supplies the required bound on the range of $f$.

For a common interval partition, the oscillation of a function on an interval satisfies ${\mathrm{osc}}_I(f+g)\le {\mathrm{osc}}_{I}f+{\mathrm{osc}}_{I}g$. Refining two good Darboux sums therefore proves integrability of the sum. Lipschitz composition preserves Riemann integrability, so absolute values preserve it too. The identity $\max (f,g)=(f+g+|f-g|)/2$ then proves the assertion for the maximum.