I have a question regarding the final step in my proof of the following exercise:
Exercise 2.34 Suppose $|f_n|\leq g \in L^1$ and $f_n \to f$ in measure. Show $\int f = \lim_{n\to\infty}\int f_n$.
Since $f_n\to f$ in measure, there exists a subsequence ${f_{n_j}}$ such that $f_{n_j}\to f$ a.e.. Now since $\forall j\in \mathbb{N}$ $|f_{n_j}|<g$ by the dominated convergence theorem we have that
$$\int f = \int \lim_{j\to\infty}f_{n_j}=\lim_{j\to\infty}\int f_{n_j}$$
Now all I need to do is show that $\lim_{j\to\infty}\int f_{n_j}=\lim_{n\to\infty}\int f_{n}$ however I cannot see why this is true.
What is the reasoning which allows those two limits to be the same?
Best Answer
This is a general fact (at least in, say, metric spaces): If you have a sequence $(f_n)$ and $f$ such that every subsequence has a further subsequence $(f_{n_{j_k}})$ converging to $f$, then the original sequence converges to $f$. This is just the negation of convergence (once you unpack the definitions). The key is that the limit is the same no matter the subsequence.
In your case, you have a sequence of real numbers $\left(\int f_n\right)_n$.