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Let $f:A\to B$ continuous such that $|f(x)-f(y)|\ge c|x-y|$ with $c>0$ constant and $x,y\in A$. Prove that, for all $g:B\to \mathbb{R}$ integrable, the composite $g\circ f:A\to \mathbb{R}$ is integrable

I'm studying analysis with a bit of measure theory. We say that a function is integrable if its Dasboux sums (inferior and superior) coincide. This answer says that it's only valid to Riemann-Lebesgue integration, which is my case, but doesn't provr anything: If $\|f(x)-f(y)\|\ge \alpha\cdot\|x-y\|$ and $g$ integrable $\Longrightarrow$ $g\circ f:A \longrightarrow \mathbb{R}$ is an integrable function

Obviously this has something to do with the oscilation of $f$, becaue of the condition $|f(x)-f(y)|\ge c|x-y|$, and remember that $w(f,X) = \sup \{|f(x)-f(y)|; x,y\in X\}$ is the max variation seen in $f$ over the set $X$. So we have a case where the oscilation is $\ge |x-y|$.

There are some theorems relating the oscilation of $f$ with its integrability. One is that for a bounded $f:A\to\mathbb{R}$ to be integrable, it's necessary and sufficient that given $\epsilon >0$, then for any partition $P$ of a block $A$ we have

$$\sum_{B\in P} w_B\cdot vol \ B <\epsilon$$

How to relate this to the composite $g\circ f$ and its integrability? I thought even about using the same criteria for the integrability of $g\circ f$, but without success.

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  • $\begingroup$ If domain of $g(x)$ is same as range of $f(x)$ and it's integrable in its domain wont it always be integrable? $\endgroup$
    – avz2611
    Oct 12, 2017 at 3:44
  • $\begingroup$ Are $A,B$ intervals? $\endgroup$
    – zhw.
    Oct 12, 2017 at 17:17
  • $\begingroup$ @avz2611 No, there are counterexamples even for smooth $f.$ $\endgroup$
    – zhw.
    Oct 12, 2017 at 18:15

3 Answers 3

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A bounded function is Riemann integrable if and only its set of discontinuity points has Lebesgue measure zero (see discontinuity theorem). So you need to prove that the set of discontinuity points of $g\circ f$ has Lebesgue measure zero. Let $$E=\{x\in A:\, g\circ f\text{ is discontinuous at }x\}$$ and $$F=\{y\in B:\, g\text{ is discontinuous at }y\}.$$ Since $f$ is continuous, if $g\circ f$ is discontinuous at $x$, then $g$ is discontinuous at $f(x)$. So $E\subset f^{-1}(F\cap f(A))$. Since $g$ is Riemann integrable, we know that $F$ has Lebesgue measure zero. On the other hand, by the property of $f$ you have that $f:A\to f(A)$ is one-to-one and onto and its inverse $h=f^{-1}$ is Lipschitz continuous. Now Lipschitz continuous functions map sets of Lebesgue measure zero into sets of Lebesgue measure zero (see point 3 in the answer to Lipschitz. Hence, $\mathcal{L}^n(h(F\cap f(A)))=0$ and in turn $\mathcal{L}^n(E)=0$.

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  • $\begingroup$ It seems that the proof for multidimensional case is similar (for instance, see this question about Lebesgue criterion. $\endgroup$ Oct 14, 2017 at 11:44
  • $\begingroup$ Also I recall that a Riemann integrable function defined on an interval is bounded (see, for instance, here). $\endgroup$ Oct 14, 2017 at 11:49
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    $\begingroup$ yes for $n\ge 2$ you still have that a bounded function is Riemann integrable iff its set of discontinuity points has Lebesgue measure zero. The proof is indeed the same. As for bounded, you only define the Riemann integrable for bounded functions, since you approximate from below and above with steps functions. For unbounded functions there is the improper or generalized Riemann integral. $\endgroup$
    – Gio67
    Oct 14, 2017 at 12:39
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Here's an elementary proof for $A,B\subset \mathbb R.$ I assume $A,B$ are closed bounded intervals.

Note that since $f$ is injective and continuous, $f$ is either strictly increasing or strictly decreasing. Let's assume the former just to form a clear picture as we move along.

Since $g$ is Riemann integrable (RI) on $B,$ it is RI on the interval $f(A)\subset B.$ Let $\epsilon>0.$ Then there exists a partition of $f(A)$ into subintervals $I_k$ such that

$$\sum_{k=1}^{n}(\sup_{I_k}g - \inf_{I_k}g)|I_k| < \epsilon.$$

Here $|I_k|$ denotes the length of $I_k.$ The last sum can be written as

$$\sum_{k=1}^{n}(\sup_{f^{-1}(I_k)}g\circ f - \inf_{f^{-1}(I_k)}g\circ f)|I_k|.$$

Now the "expansion" condition on $f$ shows that $|I_k| \ge c |{f^{-1}(I_k)}|.$ It follows that

$$\epsilon> c\sum_{k=1}^{n}(\sup_{f^{-1}(I_k)}g\circ f - \inf_{f^{-1}(I_k)}g\circ f)|{f^{-1}(I_k)}|.$$

The intervals ${f^{-1}(I_k)}$ form a partition of $A.$ Since $\epsilon$ is arbitrarily small, so is $\epsilon/c.$ Thus $g\circ f$ satisfies the Darboux criterion on $A$ as desired.

It looks to me like this proof may extend to higher dimensions, but I'm not sure.

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Previous answers give you a proof already. I just wanted to clarify a point about oscillations in the question.

For oscillations to occur, $f$ must return to the same value at different points. It can't do this because, for any $x \neq y$, $|f(x) - f(y)|> 0 \implies f(x) \neq f(y)$. This means that $f$ does not have any oscilation.

This question is simpler than you imagine. $f$ is a nice monotonic continuous invertible function. $g$ is integrable but can have discontinuities. Their composition $g \circ f$ is not necessarily continuous. The question is to show that it is integrable.

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