Does the set $\left\{\left(f,\int_a^b f\right):f\in X\right\}$ represent a function? I'm working through Undergraduate Topology by Kasriel, and the author asks the following:

Let $X$ be the set of all continuous, real-valued functions defined on
   $\left[a,b\right]$. Does \begin{align}
 \Gamma=\left\{\left(f,\int_a^bf\right):f\in X\right\},\tag{1}
 \end{align} represent a function, or further, a 1-1 function?

So far I have determined that it cannot be 1-1 because on some interval $\left[a,b\right]$, $\int_a^b f$ always results in the same value as it depends only on the endpoints. Hence, the range of this set will be the singleton $\left\{\int_a^b f\right\}$. Now, what of the domain? If $f$ is monotonic, then I'm guessing this set could be defined on all possible $\left[a,b\right]\subset\mathbb{R}$. If, however, it is periodic with period $L$, like $\cos$ or $\sin$, then it would only be a function so long as throughout the interval, some subset of that period, the function is monotonic.
Does this sound right? I feel like there's a flaw somewhere in my logic.
 A: You seem to be treating $f$ as fixed and $[a,b]$ as a variable, rather than the other way around. The function is
$$
f\mapsto \int_a^b f(x)\,dx.
$$
Does this really define a function?  In other words, given a function $f$ that is continuous on $[a,b]$, is there always a well-defined unique value of the integral?  The answer here is yes, because $f$, being continuous on a bounded interval, is bounded (so it does not have vertical asymptotes or worse) and a bounded measurable function (and continuous functions are measurable) on a bounded measurable set is integrable.
The question of whether it is one-to-one is the question of whether there could be two inputs yielding the same output:
\begin{align}
f & \mapsto \int_a^b f(x)\,dx \\[10pt]
g & \mapsto \int_a^b g(x)\,dx
\end{align}
Could both integrals be the same number even though $f$ and $g$ are different functions?  In other words, could two different functions both have the same area under their graphs between $a$ and $b$?  Draw some simple pictures and you'll see that.
A: You're misunderstanding the use of the set builder notation here. $\Gamma$ is not something that depends on a particular $f$; it is a set that (potentially) contains a pair for every possible element of $X$. We could write
$$ \textstyle \Gamma = \{ (f,\int_a^b f), (g,\int_a^b g), (h,\int_a^b h), \ldots \} $$
where $f, g, h, \ldots $ are all the various functions that are members of $X$ -- except, of course, that there are infinitely many such functions. So instead we write,
$$ \textstyle \Gamma = \{ (f,\int_a^b f) \mid f\in X \} $$
to mean, "for every function $f$ in $X$, integrate it, create the pair $(f,\int_a^b f)$ and stuff this pair into $\Gamma$".
This is a function, because if $(f,a)$ and $(f,b)$ are both in $\Gamma$ then $a$ and $b$ must be the same -- each $f$ is only used to put one pair into $\Gamma$.
Is it one-to-one? That depends on whether there can be two different functions in $X$ that both have the same integral from $a$ to $b$.
A: First of all, it is well defined since if $f=g$, then surely
$$\int_a^bf(x)\ dx = \int_a^bg(x)\ dx.$$
So it is in fact a function. You are correct in stating that it is not 1-1, in general. For example, if $a=-\pi/2,\ b=\pi/2$, then:
$$\Gamma(\sin(x))=\int_{-\pi/2}^{\pi/2}\sin(x)\ dx = 0 = \int_{-\pi/2}^{\pi/2}x^3 dx = \Gamma(x^3),$$
but $\sin(x)\neq x^3$.
