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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.