Proof that the space $L^1$ is complete in its metric

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I don't understand a particular detail of the proof on page 70 of Stein and Shakarchi's Princeton Lectures in Analysis III.

We consider a Cauchy sequence $\{f_n\}$ in $L^1$ and choose a subsequence $\{f_{n_k}\}$ such that $||f_{n_k+1} - f_{n_k}||_{L^1} \leq \frac{1}{2^k}$.

Then, we consider a function $f := f_{n_1} + \sum_{k=1}^{\infty}(f_{n_k+1}-f_{n_k})$ and observe that $|f| \leq |f_{n_1}| + \sum_{k=1}^{\infty} |f_{n_k+1} - f_{n_k}| =: g$. Everything makes sense so far.

The next steps are where I become uncertain about the specific details.

The proof then observes that

$$\int|f_{n_1}| + \sum_{k=1}^{\infty} \int|f_{n_k+1}-f_{n_k}| \leq \int|f_{n_1}| + \sum_{k=1}^{\infty}2^{-k} < \infty$$

Here is the following questions I have:

1) How is $\sum_{k=1}^{\infty} \int|f_{n_k+1}-f_{n_k}|\leq \sum_{k=1}^{\infty}2^{-k}$?

I know that $|f_{n_k+1} - f_{n_k}| \leq \frac{1}{2^k} \implies \sum_{k=1}^{\infty} \int|f_{n_k+1}-f_{n_k}| \leq \sum_{k=1}^{\infty} \frac{1}{2^k} \int d\mu$, but that begs the question, what do we do with the $\int d\mu$? In fact, often times the book is so loose with the integration symbol at times that it's not even always clear what space we're integrating over. For instance, if we're integrating over $\mathbb{R^d}$, then $\int d\mu = \infty$, so I assume this can't possibly be the case. What am I missing here?

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Your hypothesis is not that $|f_{n_k+1} - f_{n_k}| \leq \frac{1}{2^k}$, but that $$ \|f_{n_k+1} - f_{n_k}\|_1 \leq \frac{1}{2^k}. $$ That's exactly $$ \int |f_{n_k+1} - f_{n_k}| \leq \frac{1}{2^k}. $$