Suppose $X$ is a Banach space. Here, page $124$, there is this sentence
If $\mu$ is a finite support on $X$ we can define its barycenter $\beta(\mu)=\beta_X(\mu) \in X$ by $$\beta(\mu)=\int x d\mu$$
In the following Lemma $2.4$, it states that $\beta \delta = Id_X$, which I believe it to be $\beta \delta (x)=x$.
Question: How to apply the integral formula above to obtain the equation $\beta \delta = Id_X$? In other words, how to show
$$\beta(\delta(x))=\int x d\delta(x) = x$$
$\endgroup$ 11 Answer
$\begingroup$First, let me try to clarify the notation. For $x\in X$, $\delta(x)$ denotes the $\delta$ measure at the point $x$. That is, $$ \delta(x)(A) = \begin{cases} 1&x\in A\\ 0&x\notin A \end{cases} $$ for $A\subset X$ and all sets are measurable. When you integrate with respect to this measure, $x$ is a fixed parameter (which defines the measure), not a variable on integration. Your definition of $\beta$ reads $$ \beta(\mu) = \int_X y d\mu(y) $$ for a measure $\mu$, and in the case $\mu=\delta(x)$ we get $$ \beta(\delta(x)) = \int_X y d(\delta(x))(y). $$ I added parentheses for clarity and changed the variable of integration to avoid confusion. Often one would write $\delta_x$ instead of $\delta(x)$.
The integral you take is a Bochner integral since its values are supposed to be elements in a Banach space. Let us keep $x$ fixed. Any function from the measure space $(X,\delta(x))$ to $X$ is Bochner measurable and Bochner integrability of $f:X\to X$ is equivalent with $\|f(x)\|<\infty$. A simple calculation using the definitions in the Wikipedia article shows that if $f:X\to X$ is finite at $x$, then $$ \int_X f(y)d(\delta(x))(y) = f(x). \tag{*} $$ (For the sequence of simple functions $s_n$, just take $s_n(y)=f(x)$ for all $n\in\mathbb N$ and $y\in X$.)
In our case $f$ is the identity function and thus the integral defining $\beta(\delta(x))$ is well defined as a Bochner integral and evaluates to $x$.
We have thus established that $\beta(\delta(x))=x$ for all $x\in X$. Since $\delta$ is a function from $X$ to a certain space of measures on $X$ and $\beta$ is a function from that space of measures to $X$, the composition $\beta\circ\delta$ is a function $X\to X$. The result states that this is actually the identity function on $X$: $\beta\circ\delta=Id_X$. The function $\delta$ is not linear, so I would prefer to keep the circle in the composition (not writing $\beta\circ\delta$ as $\beta\delta$), but this is a matter of taste. The function $\beta$ is linear if one considers signed measures so that the space of admissible measures becomes a vector space.
Added remarks:
For any $x\in X$ our $\delta(x)$ is a measure. A measure takes sets as arguments and gives out numbers. Integration with respect to a measure (this or any other sufficiently regular measure) defines a functional from the space of continuous functions to the reals. That is, for a measure $\mu$ the functional is $f\mapsto \int_Xf(y) d\mu(y)$. Sometimes people identify a measure with the corresponding functional, but they are two different objects.
More details on (*): Compare this presentation with the Wikipedia article; the notation is introduced there. Define a sequence $(s_n)$ of simple functions on $X$ simply by $s_n(y)=f(x)$ (or, equivalently, $s_n=f(x)\chi_X$). Every function is the same constant function. We have $\int_Xs_n(y)d(\delta(x))(y)=f(x)$. The function $X\to\mathbb R$ given by $y\mapsto\|f(y)-s_n(y)\|$ takes the value $0$ at $x$. Therefore $\int_X\|f(y)-s_n(y)\|d(\delta(x))(y)=\|f(x)-s_n(x)\|=0$. (This is just the usual integral with respect to a delta measure, not a Bochner integral.) Using the definition of the Bochner integral, we have $$ \int_Xf(y)d(\delta(x))(y)=\lim_{n\to\infty}\int_Xs_n(y)d(\delta(x))(y)=\lim_{n\to\infty}f(x)=f(x). $$
Often people allow functions also to take infinite values. If the function $f$ is really a function $X\to X$ and not $X\to X\cup\{\infty\}$ for an"abstract infinity $\infty$", then of course $\|f(x)\|<\infty$ for all $x$ because all points have finite norm.
Any measure can be converted into a functional as described above. If this conversion is done for the measure $\delta(x)$, one obtains the evaluation functional at $x$. This is what equation (*) says.
