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First of all I know there are many posts where someone asked for the proof of this sentence. But I just need help to understand one step of the first part of the proof, I hope there is someone who could help me. Here is the first part of the proof:

Since f,g are integrable, they are bounded, thus f+g is bounded. Let $Z$={$x_0,...,x_n$} be a partition of [a,b]. Since $\underset{x\in [x_{j-1},x_j]} \sup (f(x)+g(x))\leq \underset{x\in [x_{j-1},x_j]} \sup f(x) +\underset{x\in [x_{j-1},x_j]} \sup g(x)$, $j=1,...,n$ we have $$U_{f+g}(Z)\leq U_f(Z)+U_g(Z)$$ (with $U_f(Z)=\displaystyle\sum_{j=1}^n$ $\underset{x\in [x_{j-1},x_j]} \sup f(x) (x_j-x_{j-1}))$. Therefore we get $$\underset{Z} \inf U_{f+g}(Z)\leq U_f(Z)+U_g(Z)$$

($\underset{Z} \inf U_{f+g}(Z)$ is the infimum of the set of all upper Darboux sums of $f+g$ regarding Z)

Let $Z_1$ and $Z_2$ be any two partitions of [a,b]. For $Z= Z_1 \cup Z_2$, we already proved that $$U_f(Z_1)\geq U_f(Z),$$ $$U_g(Z_2)\geq U_g (Z)$$ So therefore we have $$\underset{Z} \inf U_{f+g}(Z)\leq U_f(Z)+U_g(Z)\implies\underset{Z} \inf U_{f+g}(Z)\leq U_f(Z_1)+U_g(Z_2)$$ which implies that $$\underset{Z} \inf U_{f+g}(Z)\leq \underset{Z_1} \inf U_f(Z_1)+ \underset{Z_2} \inf U_g(Z_2)$$ I am able to follow this proof until this point. But then we say: This means $$\underset{Z} \inf U_{f+g}(Z)\leq \underset{Z} \inf U_f(Z)+ \underset{Z} \inf U_g(Z)$$

Why are we allowed to say this. I mean $Z_1,Z_2 \subset Z$ that means $\underset{Z} \inf U_f(Z)$ is always smaller or equal than $\underset{Z_1} \inf U_f(Z_1)$ or am I wrong. Is there anyone who could explain this conclusion for me? I would be very grateful.

Analysis_Mark
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    We call $Z$ a partition of $[a,b]$. – aschepler Apr 13 '22 at 20:10
  • Arguably, all the heavy-lifting is done in the lines before you say you are able to follow the proof up to that point. In contrast, the line you're asking about is trivially changing a variable name within the scope of a supremum (by that time, $Z, Z_1, Z_2$ aren't names for any particular partitions) – Brian Moehring Apr 13 '22 at 20:25

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This is just a renaming of variables. The number $\inf_{Z_1} U_f(Z_1)$ is the largest lower bound of all possible upper Riemann sums for $f$, and just happens to use the name $Z_1$ for any one of the possible partitions of $[a,b]$. The same goes for $\inf_Z U_{f+g}(Z)$ and $\inf_{Z_2} U_g(Z_2)$, but the names of the variables $Z$, $Z_1$, and $Z_2$ don't really matter here. At this point in the proof, those variables are no longer the particular zones with $Z = Z_1 \cup Z_2$ we were considering for a time.

So just like $\sum_{i=0}^n f(i) = \sum_{j=0}^n f(j)$ because the name of $i$ or $j$ isn't important, we have $$\inf_{Z_1} U_f(Z_1) = \inf_Z U_f(Z)$$ $$\inf_{Z_2} U_g(Z_2) = \inf_Z U_g(Z)$$

With this idea, the earlier part of the proof stating that

$$ \inf_Z U_{f+g}(Z)\leq U_f(Z)+U_g(Z) $$

is a bit sloppy or unnecessarily confusing, since on the left side $Z$ is a name for any partition at all, but on the right side $Z$ is still the particular partition introduced at the start of the proof.

aschepler
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