Bound variables in the definition of 'integrable' Spivak gives the following definition for "integrable" on p. 258 of his book "Calculus".  

A function $f$ which is bounded on [a, b] is integrable on on [a, b] if $\text{sup} \{ L(f,P):P \text{ a partition of [a, b]} \}= \text{inf} \{ U(f,P):P \text{ a partition of [a, b]} \}$

In this definition, is the partition $P$ on the left hand side of the equation necessarily the same partition $P$ on the right hand side of the equation?  I think that $P$ is a bound variable (in the sense of logic) and so the partitions may be different, but I'm not sure.  In other words, I think that this definition is equivalent:

A function $f$ which is bounded on [a, b] is integrable on on [a, b] if $\text{sup} \{ L(f,P'):P' \text{ a partition of [a, b]} \}= \text{inf} \{ U(f,P''):P'' \text{ a partition of [a, b]} \}$

Thanks in advance for your help.  Please let me know if I can clarify anything.
 A: Your interpretation is correct, i.e. you could just as well write
$$ \sup \{\, L(f,P):P \text{ a partition of }[a, b]\,\}= \inf \{\, U(f,Q):Q \text{ a partition of }[a, b]\, \}$$
A: You are correct. The scope of each $P$ is limited only to the set builder expression containing it.
The collection of all possible $P$'s is the same in both cases, of course.
A: Definitely they are bound variables and your second definition is exactly equivalent.
Sometimes people use the term "alphabetic variant" to refer to your second form.  In statement like
$$
[\,\exists x\ P(x) \,] \text{ or }[\,\forall x\ Q(x)\,]
$$
can be rewritten as
$$
[\,\exists x\ P(x) \,] \text{ or }[\,\exists y\ Q(y)\,]
$$
and that makes it possible to modify them as follows:
$$
\exists x\, \exists y\, [P(x)\text{ or }Q(y)].
$$
Alphabetic variants are used in situations like this:
$$
\begin{align}
& \sum_{i=0}^n \binom n i a^{i+1} b^{n-i} + \sum_{i=0}^n \binom n i a^i b^{n+1-i} \\[12pt]
= {} & \sum_{j=1}^{n+1} \binom n {j-1} a^j b^{n+1-j} + \sum_{i=0}^n \binom n i a^i b^{n+1-i} \tag{a substituion} \\[12pt]
= {} & \sum_{i=1}^{n+1} \binom n {i-1} a^i b^{n+1-i} + \sum_{i=0}^n \binom n i a^i b^{n+1-i} \tag{an alphabetic variant} \\[12pt]
= {} & a^{n+1} + \sum_{i=1}^n \left( \binom{n}{i-1} + \binom n i \right) a^i b^{n+1-i} + b^{n+1} \\[12pt]
= {} & a^{n+1} + \sum_{i=1}^n \binom{n+1} i a^i b^{n+1-i} + b^n+1 \\[12pt]
= {} & \sum{i=0}^{n+1} \binom{n+1} i a^i b^{n+1-i}.
\end{align}
$$
