[Math] Prove sum of two Riemann integrable functions is Riemann integrable, using approach in Analysis I of Tao

real-analysisriemann-integration

This is proposition a) in theorem 11.4.1 in Analysis I of Tao. I read through all the answers I can found in the website, but none helps to prove the proposition, because they take assumptions which are not established up to the section of the book. Can someone help?
Here is the relevant definitions and propositions provided the book.

Definition 11.3.2 (Upper and lower Riemann integrals). Let f : I → R
be a bounded function defined on a bounded interval I. We define the
upper and lower Riemann integral by the formulas. (p.c. function means piecewise constant function)

URI(f) = inf{$p.c. \int_{I} g$: g is a p.c. function on I which majorize f }.

LRI(f)= sup{$p.c. \int_{I} h$: h is a p.c. function on I which minorize f }.

Definition 11.3.4 (Riemann integral). Let f : I → R be a bounded function on a bounded interval I. f is Riemann integrable if If the upper Riemann integral on I of f is equal to the lower.

Definition 11.3.9 (Riemann sums). Let f : I → R be a bounded function on a bounded interval I, and let P be a partition of I. Upper and Lower Riemann sums,
$$U(f, P) = \sum_{J \in P; J \notin \emptyset}(sup_{x \in J} f(x)) |J|$$
$$L(f, P) = \sum_{J \in P; J \notin \emptyset}(inf_{x \in J} f(x)) |J|$$

Lemma 11.3.11. Let f : I → R be a bounded function on a bounded interval
I, and let g be a function which majorizes f and which is piecewise
constant with respect to some partition P of I. Similarly, if h is a function which minorizes f and is piecewise constant with respect to P, Then

$$p.c. \int_{I} g \ge U(f, P)$$
$$p.c. \int_{I} h \le L(f, P)$$

Proposition 11.3.12. Let f : I → R be a bounded function on a
bounded interval I. Then,

URI(f) = inf{U(f,P): P is a partition of I}

LRI(f) = sup{L(f,P): P is a partition of I}

My attempts including,

  1. Using the inequalities inf(x + y) >= inf(x) + inf(y), and the like with sup to find the inequalities among U(f+g, P), U(f, P), and U(g, P), and the like with L.
  2. Assume ($\int f + \int g) \le LRI(f+g)$, and consider a p.c. function h, which minorize $(f+g)$ and whose Riemann integral is greater than $\int f + \int g$, and tried to look for a contradiction.

Best Answer

Denote the upper and lower Riemann integrals of $f$ by $\overline \int_I f$ and $\underline \int_I f$ respectively.

We have that $U(f+g,P) \le U(f,P) + U(g,P)$ and $L(f,P)+L(g,P)\le L(f+g,P)$. Thus, $$ L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P) \le U(f,P)+U(g,P). $$

Note that the right-hand inequality gives us that, for each partition $P$, $$ \overline \int_I (f+g) \le U(f,P) + U(g,P), $$ so that $$ \overline \int_I (f+g) \le \overline \int_I f + \overline \int_I g .$$ Similarly, $$ \underline\int_I f + \underline\int_I g \le \underline\int_I (f+g). $$

We know that $f$ and $g$ are Riemann-integrable, which means that their upper and lower Riemann integrals are the same. This allows us to conclude that $$ \int_I f + \int_I g = \overline\int_I (f+g) = \underline\int_I (f+g),$$ which is what we needed to show.

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