Prove: $\overline{\int_{a}^{b}} (g+f)(x) \ dx = \int\limits_a^b {{f(x)}dx}+ \overline{\int_{a}^{b}} g(x) \ dx$

Question

I am trying to prove the following theorem:

let $f:\mathbb [a,b]\to\mathbb R$ be a Riemann integrable function, and let $g:\mathbb [a,b]\to\mathbb R$ be a bounded function.
Prove that: $$\overline{\int_{a}^{b}} (g+f)(x) \ dx = \int_a^b {{f(x)}dx}+ \overline{\int_{a}^{b}} g(x) \ dx$$

When we define:

Let $f:[a,b]\to\mathbb R$ be a function and let $P=\{a=x_0<x_1<...<x_n=b\}$ be an arbitrary partition of $[a,b]$.
$$\overline{\int_{a}^{b}} f(x) \ dx:=\inf_{P}U(f,P)$$ $$\underline{\int_{a}^{b}} f(x) \ dx:=\sup_{P}L(f,P)$$ When $U(f,P)$ is the upper Darboux sum and $L(f,P)$ is the lower Darboux sum, with respect to the partition P.

my try:

1. I have managed to prove that: $$\overline{\int_{a}^{b}} (g+f)(x) \ dx \le \overline{\int_{a}^{b}} f(x) \ dx+ \overline{\int_{a}^{b}} g(x) \ dx $$

2. Now I am trying to show that: $$\overline{\int_{a}^{b}} (g+f)(x) \ dx \ge \underline{\int_{a}^{b}} f \ dx + \overline{\int_{a}^{b}} g(x) \ dx$$

and then by using the fact that: $$ \overline{\int_{a}^{b}} f(x) \ dx = \underline{\int_{a}^{b}} f(x) \ dx \tag{*}$$

we get what we need.

The problem is that I can't figure out how to prove the second identity. I tried to play (without success) with the $L(P)$ and $U(P)$ sums and with $\sup$ and $\inf$ properties to get the result.

Any hints what to do from here?

Thanks in advance.

Answer 1

We want to show that$\newcommand{\dd}{\; \mathrm{d}}\newcommand{\ve}{\varepsilon}$ $$\overline{\int_{a}^{b}} (g+f)(x) \dd x \ge \underline{\int\limits_a^b} {{f(x)}\dd x} + \overline{\int_{a}^{b}} g(x) \dd x.$$ The question did not specify the definition of the upper/lower integral, I will assume that we're using

\begin{align*} \overline{\int\limits_a^b} f(x) \dd x &= \inf_P U(f,P)\\ \underline{\int\limits_a^b} f(x) \dd x &= \sup_P L(f,P) \end{align*}

as in the Wikipedia article on Darboux integral.

Now it suffices to notice that \begin{align*} U(f+g,P) &=\sum_{i=0}^{n+1} \sup_{t\in[x_i,x_{i+1}]}(f(t)+g(t)) (x_{i+1}-x_i)\\ &\ge\sum_{i=0}^{n+1} \left(\inf_{t\in[x_i,x_{i+1}]}f(t) + \sup_{t\in[x_i,x_{i+1}]}g(t)\right) (x_{i+1}-x_i)\\ &=\sum_{i=0}^{n+1} \inf_{t\in[x_i,x_{i+1}]}f(t) (x_{i+1}-x_i)+ \sum_{i=0}^{n+1}\sup_{t\in[x_i,x_{i+1}]}g(t) (x_{i+1}-x_i)\\ &=L(f,P)+U(g,P). \end{align*}

Now let $\ve>0$ be arbitrary. Since the lower integral is supremum of the values $L(f,P)$, there exists a partition $P_1$ such that $$L(f,P_1)\ge\underline{\int\limits_a^b} {{f(x)}dx}-\ve.$$

The same inequality is true for every partition $P$ which is finer¹ than $P_1$. For any such partition we get $$U(f+g,P) \ge U(g,P) + \underline{\int\limits_a^b} {{f(x)}dx} -\ve$$ and applying infimum to both sides we get $$\overline{\int_{a}^{b}} (g+f)(x) \dd x \ge \underline{\int\limits_a^b} {{f(x)}\dd x} + \overline{\int_{a}^{b}} g(x) \dd x-\ve.$$ Using the fact that the above is true for every $\ve>0$, we get the desired inequality $$\overline{\int_{a}^{b}} (g+f)(x) \dd x \ge \underline{\int\limits_a^b} {{f(x)}\dd x} + \overline{\int_{a}^{b}} g(x) \dd x.$$

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¹If partition $P'$ is finer than the partition $P$, then we get \begin{align*} U(f,P')&\le U(f,P)\\ L(f,P')&\ge L(f,P) \end{align*} See: Let $P'$ be a refinement of the partition $P$, then $s(f,P) \leq s(f,P') \leq S(f,P') \leq S(f,P)$, Proving the more partition points the smaller the Darboux integral gets.

²In fact, we have taken only infimum over the partitions finer than $P_1$ rather than the infimum over all partitions. But this gives the same result> If $P$ is any partition and $P'$ is a refinement of both $P$ and $P_1$, then we have $U(f,P')\le U(f,P)$. This implies that $$\inf\{U(f,P'); P'\text{ is a partition finer than }P_1\} \le \inf\{U(f,P); P'\text{ is a partition}\}.$$

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It is not the same question, but this inequality is mentioned also in the answer to this question: Additivity of definite integrals. The question can be found both using Approach Zero and using SearchOnMath.

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This might remind you of the proof of similar inequalities for limit superior and limit inferior of sequences. Indeed, with some effort we could show that the upper and lower integral can be defined as a limit superior/inferior of some net. The notion of a net is a generalization of a sequence - you could encounter such notion typically in some course on general topology. (Some references where one could find more about limit superior and limit inferior of a net are given in the answers here, here and also in the Wikipedia article on nets - I will add also a link to the current revisions.)

But even if we do not want to use nets - after all, learning about them and showing the equivalence would be non-trivial efforts - it is still useful to keep in mind that there is some kind of analogy between these two notions, so at least sometime we could expect to use somewhat similar arguments in the proofs.