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Rudin RCA p.21

Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n\}$ be a sequence of measurable functions on $X$ and suppose that

(a) $0\leq f_1\leq f_2\leq\cdots\leq\infty$ for every $x\in X,$

(b) $f_n\rightarrow f$ pointwise on $X$.

Then $\int_X f_n d\mu \rightarrow \int_X f d\mu$.


I followed the proof, but cannot see why the condition (b) is essential. Every monotonic sequence in extended real system has a limit as a supremum or infimum, so I think the condition (a) implies the condition (b). Am I wrong? How?

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You're totally right, I think (b) is just Rudin's way of labeling $f$. –  StuartHa Jan 11 '13 at 7:56
@Stuart Thank you! I was so confused.. –  Katlus Jan 11 '13 at 7:57
b) is a condition on both the sequence $f_n$ and the function $f$. –  Qiaochu Yuan Jan 11 '13 at 8:40
@QiaochuYuan: what do you mean? –  Ilya Jan 11 '13 at 8:47
I mean the condition isn't "there exists a function $f$ such that $f_n \to f$ pointwise" (which as you say is redundant), it's "$f$ is the function such that $f_n$ converges to it pointwise." The former is a condition on the sequence $f_n$ only, while the latter is a condition on a sequence $f_n$ and also another function called $f$. –  Qiaochu Yuan Jan 11 '13 at 9:26

1 Answer 1

You are right. Condition $b)$ follows from condition $a)$, but Keep in mind that we need both conditions to prove the theorem. The proof involves proving the result $\leq$ and $\geq$ to get equality. When proving \begin{equation*} \lim_{n\to\infty}\int_{\mathbb{R}^n}f_nd\mu \leq\int_{\mathbb{R}}fd\mu, \end{equation*} The monotonicity assumption $f_n\leq f_{n+1}$ implies that, for every $j$, we have \begin{equation*} f_j(x)\leq\lim_{n\to\infty}f_n(x)=f(x) \end{equation*} which is a crucial first step in the proof.

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