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My book at hand is Walter Rudin's Real and Complex Analysis, 3ed, 1987 McGraw-Hill. In Chap8: integration on product spaces, Theorem 8.16 is stated as:

Let $\mu$ be a $\sigma$-finite positive measure on some $\sigma$-algebra in some set $X$. $f:X\mapsto[0,\infty]$ be measurable.The function that assigns to each $t\in [0,\infty)$ the number $$\mu\{f>t\}=\mu(\{x\in X:f(x)>t\})$$ is called the distribution function of $f$.

8.16 Theorem Suppose that $f$ and $\mu$ are as above, that $\varphi:[0,\infty]\rightarrow[0,\infty]$ is monotonic, absolutely continuous on $[0,T]$ for every $T<\infty$, and that $\varphi(0)=0$ and $\varphi(t)\rightarrow \varphi(\infty)$ as $t\rightarrow \infty$.Then$$\int_{X}(\varphi \circ f)d\mu=\int^{\infty}_0 \mu\{f>t\}\varphi'(t)dt$$

My question is why following conditions are necessary in this theorem:

$\varphi(0)=0$ and $\varphi(t)\rightarrow \varphi(\infty)$ as $t\rightarrow \infty$

and how should I interpret the condition $\varphi(t)\rightarrow \varphi(\infty)$ as $t\rightarrow \infty$?

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  • $\begingroup$ Observe that $\infty$ is in the domain of $\varphi$. There is no room for interpretation of what $\varphi(t) \to \varphi(\infty)$ as $t \to \infty$ means but continuity at $\infty$. This is assumed to deal with functions $f$ that assume the value $\infty$. $\endgroup$
    – user66081
    Commented Feb 12, 2015 at 17:03
  • $\begingroup$ @user66081 So it just mean that this function is continuous at infinity? But why is that necessary to this theorem? I mean is there example that this theorem fails when this condition is dropped? $\endgroup$
    – Henry.L
    Commented Feb 12, 2015 at 17:04
  • $\begingroup$ Ok; suppose $f \equiv \infty$, and suppose $\varphi(t) := t$ for $t > 0$ and $\varphi(\infty) := 0$. Then the left-hand-side equals zero, while the right-hand-side does not (unless the measure of $X$ is zero, but maybe this is excluded in ``$\mu$ is a positive measure''). Agree? $\endgroup$
    – user66081
    Commented Feb 12, 2015 at 17:06
  • $\begingroup$ @user66081 Now I am confused with your example. how could such a $/varphi$ be continuous at infinity? $\endgroup$
    – Henry.L
    Commented Feb 12, 2015 at 17:10
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    $\begingroup$ You asked for an example where the theorem does not hold when this condition (of continuity at $\infty$) is dropped; that was an example $\endgroup$
    – user66081
    Commented Feb 12, 2015 at 17:11

2 Answers 2

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The theorem need not hold if one of the conditions a) $\varphi(0) = 0$, b) $\varphi$ is continuous at $\infty$, is dropped:

Ad a): Take a smooth $\varphi$ with $\varphi(0) \neq 0$, and $f \equiv 0$. Then the left-hand-side is nonzero (since $\mu(X) \neq 0$), but the right-hand-side is zero.

Ad b): Suppose $f \equiv \infty$, and suppose $\varphi(t) := t$ for $t > 0$ and $\varphi(\infty) := 0$. Then the left-hand-side equals zero, while the right-hand-side does not (since $\mu(X) \neq 0$).

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  • $\begingroup$ Sure. Thanks a lot. $\endgroup$
    – Henry.L
    Commented Feb 12, 2015 at 17:19
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Assuming you copied that correctly, there is no reason why $\phi$ should tend to $\infty$ as $t$ does. Think of $\arctan$ as an example. The assumption, as stated, is simply that $\phi(t)$ has a well defined limit as $t\rightarrow \infty$.

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  • $\begingroup$ Is says more: the limit equals $\varphi(\infty)$. $\endgroup$
    – user66081
    Commented Feb 12, 2015 at 17:03
  • $\begingroup$ So why is this assumption necessary? I checked again and I am sure that I have copied it correctly. $\endgroup$
    – Henry.L
    Commented Feb 12, 2015 at 17:06

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