Why is the marginal liklihood term(the denominator) in Bayes' updation not a function of theta(parameters)? 
The snippet from the book is pasted above. I don't understand how the denominator is not a function of theta as it is integrated over theta. Thank you.
 A: When you calculate a definite integral over some variable, and the limits don't contain that variable, the variable doesn't appear in the value. Indeed, you can substitute a different variable (integrate w.r.t. a different dummy) while leaving the integral unchanged.
So 
$$\int_{-\infty}^{\infty} g(\theta) d\theta$$ 
isn't a function of $\theta$.

Edit - since more explanation is required (I am trying to convey an intuitive understanding, so this is a bit handwavy):
Note that when you do an indefinite integral, $\int g(x) dx$, the variable can appear in the result: $G(x)+c$. But when you do a definite integral $\int_a^b g(x) dx$, you substitute the limits into the antiderivative $(G(x)+c)|_a^b$, so every place the variable appeared in the antiderivative is replaced by the limits: $G(b)-G(a)$, leaving the variable absent from the result. 
[This is not restricted to linear functions!]
A: One source of confusion here may be that $\theta$ is used to denote two different things in the equation. $\theta$ appearing in the numerator is the same $\theta$ as in the left-hand-side, namely, the particular parameter values whose posterior density is obtained by substituting that particular $\theta$ on the right-hand-side. 
Meanwhile, the $\theta$ in the denominator is just the variable of integration. As pointed out by @Glen_b, a definite integral is not a function of the variable of integration. Especially,
\begin{equation}
\int f(\alpha) d\alpha = \int f(\theta) d\theta ,
\end{equation}
as long as the limits (the area over which the integral is computed) are the same. It might be beneficial to write Bayes' rule using a different symbol in the denominator, but I guess the same symbol is used to denote that both refer to the same quantity, i.e., the same parameters of the model.
An example
Consider tossing a (possibly biased) coin once, and observing heads. Let our parameter $\theta$ be the probability that the coin comes up heads on a single toss. Define prior (before the toss) as uniform
\begin{equation}
p(\theta) = 1,~\theta \in [0,1].
\end{equation}
Now, by definition of the parameter (and our data: heads once), 
\begin{equation}
p(D\mid \theta) = p(\mathrm{heads} \mid \theta) = \theta
\end{equation}
Thus, the numerator is $p(\theta)p(D \mid \theta) = \theta$ when $\theta \in [0,1]$. The denominator is 
\begin{equation}
\int_0^1 p(\theta) p(\mathrm{heads} \mid \theta) d\theta = \int_0^1 \theta d\theta = \left. \frac{1}{2}\theta^2 \right|_{\theta=0}^{\theta = 1} = \frac{1}{2}.
\end{equation}
As expected, this is not a function of $\theta$. It would be a function of the data $D$, but here for simplicity I have considered only a fixed $D$. Note that we could have denoted the parameter by any other symbol while computing the denominator:
\begin{equation}
\int_0^1 p(\alpha) p(\mathrm{heads} \mid \alpha) d\alpha = \int_0^1 \alpha d\alpha = \left. \frac{1}{2}\alpha^2 \right|_{\alpha=0}^{\alpha = 1} = \frac{1}{2}.
\end{equation}
The posterior is then
\begin{equation}
p(\theta \mid D=\mathrm{heads}) = \frac{p(\theta)p(\mathrm{heads} \mid \theta)}{\int_0^1 p(\theta) p(\mathrm{heads} \mid \theta) d\theta} =  \frac{\theta}{\frac{1}{2}} = 2\theta,
\end{equation}
which is indeed proportional to $p(\theta)p(\mathrm{heads} \mid \theta)=\theta$.
A: The denominator in 11.1 is a function of $\theta$. However the denominator takes the same value irrespective of of which particular value of $\theta$ we are interested in because we intergrate over the whole range of possible values for $\theta$. The equation below 11.1 states that $P(\theta|D)$ is proportional to $P(\theta)L(\theta;D)$, not equal to it. i.e. 
$$P(\theta|D) = kP(\theta)P(D|\theta)$$
Here, the constant of proportionality, $k$ is the denominator form 11.1:
$$ k= \frac{1}{\int{P(\theta)P(D|\theta)d\theta}} $$
We can use this trick because often we are interested either in finding the $\theta$ that maximizes $P(\theta|D)$ rather than the value of $P(\theta|D)$ and so the constant doesn't matter, or we are interested in the ratio of $P(\theta|D)$ for two different values of theta, and so the $k$s cancel out. 
This is a useful trick because the denominator can be difficult or impossible to calculate in many cases. 
