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?

  • 7
    $\begingroup$ You're totally right, I think (b) is just Rudin's way of labeling $f$. $\endgroup$ Jan 11, 2013 at 7:56
  • $\begingroup$ @Stuart Thank you! I was so confused.. $\endgroup$
    – Katlus
    Jan 11, 2013 at 7:57
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    $\begingroup$ b) is a condition on both the sequence $f_n$ and the function $f$. $\endgroup$ Jan 11, 2013 at 8:40
  • $\begingroup$ @QiaochuYuan: what do you mean? $\endgroup$
    – SBF
    Jan 11, 2013 at 8:47
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    $\begingroup$ 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 $f$. $\endgroup$ Jan 11, 2013 at 9:26

2 Answers 2


If you omitted condition (b), then nothing in the hypothesis tells you what $f$ is.

Remember that a theorem should be correct no matter what particular values you give its variables; as long as the hypotheses are true, the conclusion must be true also. Now suppose you chose some reasonable functions as your $f_n$'s, satisfying hypothesis (a), so they converge, but you chose some totally different function as your $f$, not the limit to which the $f_n$'s converge. For this choice of $f_n$'s and $f$, the conclusion would probably be false (unless you happened to choose a particularly lucky $f$), but all the hypotheses except (b) are true.

Therefore, if you omit (b), the theorem becomes incorrect. There are choices of $f_n$'s and $f$ that make the surviving hypotheses true but make the conclusion false.

It is possible to omit hypothesis (b) and compensate for the omission so as to keep the theorem correct. For example, by writing the last integral in the conclusion as $$\int_X\lim_{n\to\infty}f_n\,d\mu.$$


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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