I am studying the measure theory, and I am doing some exercises. Here is an exercise question. I have got the answer for it but I am confused about the answer. Let me first state the question, then give the proof from the solution, and then I will point out where I am confused.
Question:
Suppose $\{f_{n}\}$ is a sequence of real-valued functions in integrable on a set $A$ of finite measure. Show that if the sequence is uniformly convergent on $A$, then $$\lim_{n\rightarrow\infty}\int_{A}f_{n}=\int_{A}\lim_{n\rightarrow\infty}f_{n}$$
Proof:
Since $\{f_{n}\}$ is uniformly convergent on $A$, there is $n_{0}$ such that $|f_{n}(x)-f_{n_{0}}(x)|\leq 1$ for $n>n_{0}$ and $x\in A$. Thus $|f_{n}|\leq|f_{n_{0}}|+1$, and since $A$ is of the finite measure, $|f_{n_{0}}|+1$ is integrable on $A$. So we can apply Lebesgue Dominated Convergence Theorem.
My Questions:
1) In the proof, he used a definition of uniform convergence that I do not quite understand. Usually, if I am not wrong, we use the definition: for all $\epsilon>0$, $\exists N=N(\epsilon)$, such that $|f_{n}(x)-f(x)|<\epsilon$ for all $n>N$ and $x\in A$. So my question here is that, how and why does the solution come out the $n_{0}$ in $f_{n_{0}}(x)$, why he used $1$ on the right side instead of $\epsilon$?
2) Even though we assume the first inequality --- $|f_{n}(x)-f_{n_{0}}(x)|\leq 1$, is right. Why does this inequality imply $|f_{n}|\leq|f_{n_{0}}|+1$?
3) Finally, why does the fact that $A$ is of finite measure can guarantee $|f_{n_{0}}|+1$ is integrable? For a non-negative function, we need pretty many other criterions to ensure it is integrable, right?
I don't know whether I clearly explain my questions, so please do not hesitate to point out the ambiguity in my descriptions.
Thank you!
Edit 1: {Lebesgue Dominated Convergence Theorem}
Suppose $\{f_{n}\}$ is a sequence of measurable functions on $A$ and $f(x)=\lim_{n\rightarrow\infty}f_{n}(x), x\in A$. If there exists a function $g$ integrable on $A$ such that $|f_{n}(x)|\leq g(x)$ for all $x$ and $n$, then we have $$\lim_{n\rightarrow\infty}\int_{A}f_{n}=\int_{A}f$$
Since the sequence is uniformly convergent in $A$, it is uniformly Cauchy in $A$, which means "$\forall\epsilon>0$ $\exists n_0\in\mathbb{N}$ such that $|f_m(x)-f_n(x)|<\epsilon$ whenever $m,n\geq n_0$ for all $x\in A$". In particular, letting $\epsilon=1$ and $m=n_0$ we get the desired statement.
It is just triangle inequality: for $a,b\in\mathbb{R}$ we know that $$|a|-|b|\leq |a-b|$$
By hypothesis the elements of $\{f_n\}$ are integrable on $A$, thus $f_{n_0}$ is integrable and $1$ is also integrable on $A$ (here we are using finite measure), therefore $|f_{n_0}|+1$ is integrable on $A$.