Sampling Methods : Pattern Recognition and Machine Learning Bishop

I am reading chapter 11 . Sampling Methods from the book : Pattern Recognition and Machine Learning by Bishop : In the introduction , in short,he evaluates expectation of some function $$f(z)$$ with respect to a probability distribution $$P(z)$$ where $$z$$ is the random variable . He writes :
\begin{align} E(f) = \int f(z)p(z)dz \end{align}
Now he says that we suppose that such expectations are too complex to be evaluated exactly using analytical techniques . So we use sampling. And the idea behind sampling methods is to obtain a set of samples $$z^{(l)}$$, (where $$l=1,....,L$$) drawn independently from the distribution $$P(z)$$. This allows the expectation to be written as \begin{align} f^{\hat{}} = \frac{\sum_{l=1}^{L}f(z^{(l)})}{L} \end{align} Now further in the official solution for excercise 11.1 , he calculates : \begin{align} E[f^{\hat{}}] = \frac{1}{L}\sum_{l=1}^{L}\int f(z^{(l)}) p(z^{(l)})dz^{(l)}\tag 1 \end{align} Now i did not understand this integral .
My argument , for simplicity , let us assume the underlying distribution $$P(z)$$ to be a one dimensional standard normal distribtion .
1.) When the author samples $$z^{1}$$, i assume that he drew a fixed $$n$$ values which would look something like : $$z^{1}= \{0.2,0.30.2323,... ,0.8\}$$ ,total $$n$$ values .
Now in that case how would the integeral in $$(1)$$ look like ? For example what would be the limits of integration ?

If $$P(z)$$ is a one-dimensional normal distribution, then there is no $$n$$. Each sample $$z^{(l)}$$ is a single draw from $$P(z)$$, i.e. $$z^{(l)}$$ is just a real number. The integral is over the support of the distribution. In this case, the limits of integration are $$(-\infty, \infty)$$.