Let $[a,b]$ be a compact interval, and let $F: [a,b] \rightarrow {\bf R}$ be a continuous function. Then one can find an at most countable family of disjoint non-empty open intervals $I_n = (a_n,b_n)$ in $[a,b]$ with the following properties: \begin{enumerate}[label=(\roman*)] \item For each $n$, either $F(a_n)=F(b_n)$, or else $a_n=a$ and $F(b_n) \geq F(a_n)$. \item If $x \in (a,b]$ does not lie in any of the intervals $I_n$, then one must have $F(y) \leq F(x)$ for all $x \leq y \leq b$. \end{enumerate}

Homepage › Solution manuals › Terence Tao › An Introduction to Measure Theory › Lemma 1.6.17 (Rising sun lemma)

Lemma 1.6.17 (Rising sun lemma)

Let $[a, b]$ be a compact interval, and let $F : [a, b] \to R$ be a continuous function. Then one can find an at most countable family of disjoint non-empty open intervals $I_{n} = (a_{n}, b_{n})$ in $[a, b]$ with the following properties:

(i): For each $n$ , either $F (a_{n}) = F (b_{n})$ , or else $a_{n} = a$ and $F (b_{n}) \geq F (a_{n})$ .
(ii): If $x \in (a, b]$ does not lie in any of the intervals $I_{n}$ , then one must have $F (y) \leq F (x)$ for all $x \leq y \leq b$ .

Lemma 1.6.17 (Rising sun lemma)

Add answer