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I am trying to figure out the following problem in measure theory and am stuck. It seems like it should be very easy, so I must be missing something.

Let $g: \mathbb{R} \to \mathbb{R}$ be a mapping of $\mathbb{R}$ onto $\mathbb{R}$ for which there is a constant $c > 0$ for which

$$ |g(u) - g(v)| \geq c \cdot |u-v| \text{ for all } u, v \in \mathbb{R}. $$

(Note to avoid confusion: this function is NOT Lipschitz and not supposed to be.)

Show that if $f: \mathbb{R} \to \mathbb{R}$ is Lebesgue measurable, then so is the composition $f \circ g$.

I see that we need to show that $g$ maps measurable sets to measurable sets. I know how to show $g$ is injective and that bounded sets are mapped to and from bounded sets... but I'm not sure where to go from there.

I'd appreciate a nudge in the right direction. Please do not give away the whole problem, if possible.

t.b.
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Bill Karr
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    Are you sure you don't want Lipschitz continuity, that is $|g(u) - g(v)| \leq c|u-v|$? Moreover, you don't want to check that $g$ maps measurable sets to measurable sets, you want that preimages of measurable sets are measurable. For this (assuming my guess in the first sentence is right), recall that every measurable set is a union of a Borel set and a null set, so by continuity of $g$ it suffices to show that the pre-image of a null-set is a null-set. – t.b. Apr 19 '11 at 01:39
  • Yes, I am sure it is not Lipschitz. This is written correctly... it's number 27 in chapter 20 in Royden's 4th edition book. We proved this for Lipschitz continuous functions a long time ago.

    Also, you can show that if $g$ has the above condition, it is one to one, so mapping measurable sets to and from measurable sets are the same thing.

     – Bill Karr Apr 19 '11 at 01:41
  • If it helps at all, this problem is in the section that goes over Lebesgue measure on $\mathbb{R}^n$ and the composition rule for integration when you compose an integrable function with an invertible linear map. – Bill Karr Apr 19 '11 at 01:43
  • Ah, ok. I should have paid closer attention to what you wrote afterwards. Sorry for the noise. So if you look at images of null sets under $g$, they will most certainly be null, I guess, but what I don't see is why the image of a Borel set is still measurable (if we assume $g$ to be Borel, I can prove this, but this is probably an overkill argument). – t.b. Apr 19 '11 at 01:51
  • Yes, this problem seems kind of out of nowhere and unrelated to what we are doing... but usually that just means I'm not interpreting it in terms of the stuff we're doing right now. I think the onto assumption is crucial. I have already used it to show bounded sets go to bounded sets... but I'm not sure where to go after that. – Bill Karr Apr 19 '11 at 02:04
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    My measure theory is rusty, but since $g$ is 1-1, doesn't it follow that $g^{-1}$ is Lipscitz? Then you should get it from there... – N. S. Apr 19 '11 at 02:07
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    Ah, I overlooked the onto part. So $g$ has a Lipschitz continuous inverse, but that means that $g$ maps Borels to Borels and nulls to nulls (hope I didn't give too much away like that). – t.b. Apr 19 '11 at 02:08
  • Hahaha.. I can't believe I didn't see that. It's okay! That is why I said "if possible." Thank you very much! – Bill Karr Apr 19 '11 at 02:31
  • Well, we sort of nudged each other through this argument, didn't we? Cool problem, didn't know that! – t.b. Apr 19 '11 at 02:34
  • Definitely! I am a new user, so I'm not sure how to mark this question answered. Do you have to post an official answer or something? – Bill Karr Apr 19 '11 at 02:36
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    Well, I suggest that you type up a short account of the argument and then accept it (I need to go to bed now). By the way: add a twitter-like @user if you want to notify someone, otherwise they don't see that you've answered them. – t.b. Apr 19 '11 at 02:40
  • Okay, I will tomorrow. I joined this website today, and it won't let me answer my own questions, yet. – Bill Karr Apr 19 '11 at 02:51
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    @Bill: what say you write up that short account of the argument that was hashed out between you and @Theo? – Willie Wong Jul 30 '11 at 21:25
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    @Willie: done, after a long time. Sorry for the delay... – t.b. Dec 25 '11 at 11:45

1 Answers1

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The point here is that the assumptions that $g: \mathbb{R} \to \mathbb{R}$ is surjective and that $|g(x)-g(y)| \geq c|x-y|$ for some $c \gt 0$ imply that $g$ is bijective and that its inverse $h: \mathbb{R} \to\mathbb{R}$ is Lipschitz continuous with Lipschitz constant $c^{-1}$.

In particular, $h$ maps Lebesgue measurable sets to Lebesgue measurable sets. In more detail:

  1. $h$ maps Borel sets to Borel sets: Every closed set is a countable union of compact sets and thus $h$ maps closed sets to Borel sets because $h(\bigcup C_n) = \bigcup h(C_n)$ where the right hand side is a countable union of compact sets by continuity of $h$, hence it is Borel measurable. This tells us that $g = h^{-1}$ is Borel measurable, so $h$ indeed maps Borel sets to Borel sets.(*)
  2. $h$ maps null sets to null sets: If $N$ is a null set then given any $\varepsilon \gt 0$ we can cover $N$ by countably many intervals whose total length doesn't exceed $\varepsilon$. By Lipschitz continuity, the images of these intervals will again be intervals whose total length doesn't exceed $c^{-1} \varepsilon$, and this shows that $h(N)$ is a null set.
  3. Every Lebesgue measurable set $L$ can be written as $L = B \cup N$ with $B$ Borel and $N$ null. Then $h(L) = h(B) \cup h(N)$ is a union of a Borel set and a null set by 1. and 2., so $h(L)$ is Lebesgue measurable.

In conclusion, $g^{-1}(L) = h(L)$ is Lebesgue measurable for every Lebesgue measurable $L$. Thus, $g^{-1}(f^{-1}(B))$ is Lebesgue measurable for every Borel set $B$ because $L = f^{-1}(B)$ is Lebesgue measurable by measurability of $f$, and this proves that $f \circ g$ is measurable.

(*) It is a general fact due to Lusin and Souslin that a Borel measurable injection between completely metrizable spaces sends Borel sets to Borel sets, but this is much more difficult to prove than 1 above (see e.g. Kechris, Classical descriptive set theory, Theorem 15.1, page 89). Without injectivity this fails, the continuous image of a Borel set is not a Borel set: it is only analytic in general. See this MO answer for the standard story to be told at this point…

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