To directly answer your question of "is this step necessary?", the answer is it depends. It depends on whether or not you actually know how to prove that it is equivalent to the usual statement which only has $<\epsilon$ at the end. If you already know the equivalence, then of course you don't have to keep repeating it. If you don't know the equivalence, then of course, you have to go the extra step; after all, the reason why sometimes we might go the extra step of starting with $\frac{\epsilon}{2(b-a)}$ or starting with a different quantitity $\epsilon_1$ is just to make contact directly with the theorems proved/definitions involved.
Perhaps that last part may have been a little cryptic, so let me instead illustrate with an example. Suppose $f:I\to \Bbb{R}$ is a given function where $I$ is say an open interval containing the point $a$. Also, let $l\in\Bbb{R}$. Consider now the following statements:
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|<\delta$ then $|f(x)-l|<\epsilon$.
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|<\delta$ then $|f(x)-l|\leq \epsilon$.
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|\leq \delta$ then $|f(x)-l|< \epsilon$.
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|\leq\delta$ then $|f(x)-l|\leq \epsilon$.
In other words, I have just replaced some of the strict inequalities with weak inequalities. Now consider the next "type" of statement:
- there is a $k>0$ such that for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|<\delta$ then $|f(x)-l|< k\epsilon$.
- there is a $k>0$ such that for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|<\delta$ then $|f(x)-l|\leq k\epsilon$.
- there is a $k>0$ such that for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|\leq\delta$ then $|f(x)-l|< k\epsilon$.
- there is a $k>0$ such that for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|\leq\delta$ then $|f(x)-l|\leq k\epsilon$.
So, I have added the prefix "there exists $k>0$" and finished off with $k\epsilon$, again with the four variants of weak/strict inequalities. Finally, let me introduce the following type of statement. Suppose $\psi:(0,\infty)\to (0,\infty)$ is a function such that $\lim\limits_{\epsilon\to 0^+}\psi(\epsilon)=0$. Now consider the following statements:
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|<\delta$ then $|f(x)-l|<\psi(\epsilon)$.
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|<\delta$ then $|f(x)-l|\leq\psi(\epsilon)$.
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|\leq\delta$ then $|f(x)-l|<\psi(\epsilon)$.
- for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0<|x-a|\leq\delta$ then $|f(x)-l|\leq \psi(\epsilon)$.
So, the twelve statements I made above, with various choices of weak/strict inequalities, and with various choices of the final quantity ($\epsilon$ vs $k\epsilon$ vs $\psi(\epsilon)$), it turns out they are all logically equivalent. To someone who has finished a first course in real analysis, these equivalences tend to become almost obvious (furthermore, here I only stated things for functions defined on subsets of $\Bbb{R}$ and with values in $\Bbb{R}$, but the same can be generalized to metric spaces and stuff). The key to understanding the equivalence is the role of the quantifiers "for every" and "there exist".
Point ($1$) is often taken as the definition of "$\lim\limits_{x\to a}f(x)=l$", so if you start with that as the definition, you have to prove the remaining statements as theorems. Often times textbooks avoid mentioning that these statements are all equivalent (some textbooks may leave some of these equivalences as exercises, such as Spivak). Also, since point (1) is the definition straight out of a textbook, often in an attempt to not to invoke results not already proved, they may resort to making their proofs slightly more opaque by bringing in $\frac{\epsilon}{2}$ or $\frac{\epsilon}{2k}$ or stuff like that; in a sense this method of brining in "auxillary epsilons" is pretty much a way of proving the equivalence of the above statements, without explicitly saying that's what they're doing.
For the case of the upper and lower sums, it's the same story. The statement
- for every $\epsilon>0$ there is a partition $P$ of the interval $[a,b]$ such that $U(f,P)-L(f,p)<\epsilon$
can also be expressed equivalently in many ways. We can use either $<$ or $\leq$
at the end. We can add the prefix "there is a $k>0$" in the beginning and finish off with $<k\epsilon$; we can also suppose there is a function $\psi$ as described previously and end with $<\psi(\epsilon)$.
So once again, the question of "is this step necessary" depends entirely on what you have/have not already proved. From a strict logical perspective, yes if you start with one of the statements as a definition, you have to prove that the remaining eleven are equivalent. However, to many "experienced" people (where "experienced" means usually by the end of a 1-2 semester course in analysis) these equivalences are almost obvious, so going the extra step may be unnecessary.
In practice, we often end up proving something like statement (8) (where usually we have some explicit value for $k$ like $k=2$ or $k=10$ or $k=2\sup\limits_{x\in [a,b]}|f(x)|$) or (12) instead, because weak inequalities are more forgiving, and that's all we really need.
Best Answer
You can express $\max\{ f,g \}$ as $$ \frac{f+g + |f-g|}{2} $$ Then $\sup$ of the $\max$ would be the $\sup$ of this function. However, better than resorting to the $\epsilon$ definition is to prove that $\frac{f+g+|f-g|}{2}$ is integrable whenever $f$ and $g$ are. You could do this by proving that the sum (and difference) of integrable functions is integrable and that the absolute value of an integrable function is integrable.