How to perform abstract integration 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
$$\beta(\delta(x))=\int x d\delta(x) = x$$
 A: 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.
