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This is from P.580 of Serge Lang's undergraduate analysis (2nd edition).

$\textbf {Proposition 2.3.}$ Let $A$ be an admissible set in $\mathbb R^n$ and assume that its closure $\bar{A}$ is contained in an open set $U$. Let $f \colon U \to \mathbb R^n$ be a $C^1$ map, which is $C^1$-invertible on the interior of $A$. Then $f(A)$ is admissible and $$\partial f(A)\subset f(\partial A).$$

Proof. Let $A^o$ be the interior of $A$, that is the set of points of $A$ which are not boundary points of $A$. Then $A^o$ is open, so is $f(A^o)$, and $f$ yields a $C^1$-invertible map between $A^o$ and $f(A^o)$. We have $$\bar{A}=A^o \cup \partial A,$$ and $\partial A = \partial \bar{A}$, whence $$f(A^o) \subset f(A) \subset f(\bar{A}) = f(A^o) \cup f(\partial A).$$ This shows that $\partial f(A)\subset f(\partial A),$ and that $\partial f(A)$ is negligible by Proposition 2.1, thus proving Proposition 2.3.

I have difficulty in understanding the proof, hope that someone here can enlighten me.

My questions are:

i) On line 4 of the proof, why is it true that $\partial A = \partial \bar{A}$?

ii) Why do we need this fact in this proof?

I tried to ask the questions in MSE but received no reply. Thank you in advance for taking time to answer my question.

PS: According to Lang, a set is said to be admissible if it is bounded and its boundary is a negligible set. This is usually called a Jordan domain. A negligible set is merely a set of Jordan content zero.

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    $\begingroup$ Hush-hush, don't use word undergraduate around here. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Oct 20, 2014 at 5:41
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    $\begingroup$ So if $A=(-1,1)\setminus\{0\}$, then it's a bounded subset of $\mathbb R^1$ and its boundary consists of $\{-1,0,1\}$. On the other hand $\bar A=[-1,1]$, so that its boundary consists of $\{\pm1\}$. Hence, as you suspected it's not true in general that $\partial A=\partial\bar A$. Also, as you suggest, I see no place where the supposed equality of these two sets is used. $\endgroup$
    – Anthony Quas
    Commented Oct 20, 2014 at 5:57
  • $\begingroup$ The MSE post is here: math.stackexchange.com/questions/981235/… Despite the slightly sarcastic comment by Włodzimierz, I do feel that MSE is a more appropriate site for this question. $\endgroup$
    – Todd Trimble
    Commented Dec 19, 2014 at 13:01

1 Answer 1

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Since $\bar A=\mathring A\cup \partial A$, it holds that $f(\mathring A)\subset f(A) \subset f(\bar A)=f(\mathring A)\cup f(\partial A)$.

Because $f$ is $C^1$-invertible on $\mathring A$, the inverse function theorem implies that $f(\mathring A)$ is open in $\mathbf R^n$. Moreover, $f$ is continuous on $U$ and $\bar A$ is compact (bounded and closed), hence $f(\bar A)$ is compact. This implies that $f(\mathring A)\subset f(A)^\circ \subset f(A)\subset f(\bar A)$, and $\partial f(A) \subset \partial f(\bar A)\subset f(\partial A)$.

Since $f$ is $C^1$ on $U$ and $\partial A$ is negligible, then $f(\partial A)$ is negligible too. This implies that $f(A)$ is negligible.

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