I was reading Rudin's Principles of Mathematical Analysis, and had a question about the Remark 11.23(e) on page 315:

(e) If $\mu(E)=0$, and $f$ is measurable, then $\int_E fd\mu=0.$

Is this remark still true if $f(x)=\infty$ on $E$? If so, how do we justify it? like the following?

Recall that $\int_E fd\mu\triangleq \underset{s}\sup I_E(s)$, where $s$ is a simple function, $s(x)=\sum_{i=1}^{n}c_iK_{E_i}(x)$, satisfying $0\le s\le f,$ and $I_E(s)=\sum_{i=1}^{n}c_i\mu(E\cap E_i).$ By definition (page 313), a simple function is a real-valued function with a finite range. I assume that this means $s$ is not extended real-valued, i.e. it can't take value of $\infty$. With this assumption, then, clearly $I_E(s)=0$ for any simple function $s$. As a result, $\underset{s}\sup I_E(s)=0.$

So it appears that $s$ being real-valued, but not extended real-valued, is key for $\int_E \infty d\mu=0$ to hold. Is it correct?


  1. This question is related to the following post (Integral over set of measure zero.), but I want to emphasize and clarify the justification when $f(x)=\infty$ on $E$.

  2. According to Rudin's book (definition 11.13, page 310), a measurable function is an extended real-valued function:

    11.13 Definition Let $f$ be a function defined on the measurable space $X$, with values in the extended real number system. The function $f$ is said to be measurable if the set $\{x\mid f(x)>a\}$ is measurable for every real $a$.

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    $\begingroup$ In measure theory we take the general convention that $\infty\cdot 0=0$. So if you approximate $f=\infty$ by the simple function function $f_n = n$ then $f_n\rightarrow f$ but $\int_E f_nd\mu = 0$ for all $n$. $\endgroup$ – TSF May 10 '17 at 2:43
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    $\begingroup$ We take the convention $\infty\cdot 0=0$ in part because this result is true, not the other way around. I don't understand the "So" @TonyS.F. $\endgroup$ – Jonas Meyer May 10 '17 at 3:39
  • $\begingroup$ $\infty\cdot 0$ is not always $0$. This is trivial, think of $\lim\limits_{n\rightarrow\infty}1 = \lim\limits_{n\rightarrow\infty}n\frac{1}{n} = \infty\cdot 0 =1$. $\infty\cdot 0=0$ is something you accept for measure theory but that does not hold in general. We don't calculate this integral and then decide to make $\infty\cdot 0=0$, we have to accept that before hand. $\endgroup$ – TSF May 10 '17 at 6:20
  • $\begingroup$ @TonyS.F.: Is this a response to my comment? My point is that the convention is useful when it reflects relevant facts, but the convention is not a reason anything is true. You say "So if...". Usually "So..." after making a statement indicates the previous statement helps explain why something is true. "We have to accept that before hand." No, the integral has a definition and with this definition it comes out to $0$. The convention merely organizes some true statements about integrals. $\endgroup$ – Jonas Meyer May 10 '17 at 6:23

We use the monotone convergence theorem to show this. Note that $f_n=n\chi_{E}$ is a sequence of functions which converge monotonically to $f=\infty$ on $E$. Then, $\int fd\mu = \lim\int f_nd\mu = \lim n\mu(E) = \lim n\cdot 0= 0$.

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    $\begingroup$ Those equations hold already from the definition of the integral without having to apply MCT. $\endgroup$ – Jonas Meyer May 10 '17 at 3:38
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    $\begingroup$ You're right, just apply the definition of integral of a nonnegative function. Any simple function will have integral $0$ on $E$ so the sup will be $0$ as well. $\endgroup$ – TSF May 10 '17 at 6:25

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