118 1. Measure theory (i) For each n, either F (an) = F (bn), or else an = a and F (bn) ≥ F (an). (ii) If x ∈ [a, b] does not lie in any of the intervals In, then one must have F (y) ≤ F (x) for all x ≤ y ≤ b. Remark 1.6.18. To explain the name “rising sun lemma”, imagine the graph {(x, F (x)) : x ∈ [a, b]} of F as depicting a hilly landscape, with the sun shining horizontally from the rightward infinity (+∞, 0) (or rising from the east, if you will). Those x for which F (y) ≤ F (x) are the locations on the landscape which are illuminated by the sun. The intervals In then represent the portions of the landscape that are in shadow. The reader is encouraged to draw a picture14 to illustrate this perspective. This lemma is proven using the following basic fact: Exercise 1.6.10. Show that any open subset U of R can be written as the union of at most countably many disjoint non-empty open intervals, whose endpoints lie outside of U. (Hint: First show that every x in U is contained in a maximal open subinterval (a, b) of U, and that these maximal open subintervals are disjoint, with each such interval containing at least one rational number.) Proof. (Proof of rising sun lemma) Let U be the set of all x ∈ (a, b) such that F (y) F (x) for at least one x y b. As F is continuous, U is open, and so U is the union of at most countably many disjoint non-empty open intervals In = (an,bn), with the endpoints an,bn lying outside of U. The second conclusion of the rising sun lemma is clear from construction, so it suﬃces to establish the first. Suppose first that In = (an,bn) is such that an = a. As the endpoint an does not lie in U, we must have F (y) ≤ F (an) for all an ≤ y ≤ b similarly, we have F (y) ≤ F (bn) for all bn ≤ y ≤ b. In particular, we have F (bn) ≤ F (an). By the continuity of F , it will then suﬃce to show that F (bn) ≥ F (t) for all an t bn. Suppose for contradiction that there was an t bn with F (bn) F (t). Let A := {s ∈ [t, b] : F (s) ≥ F (t)}, then A is a closed set that contains t but not b. Set t∗ := sup(A), then t∗ ∈ [t, b) ⊂ In ⊂ U, and thus there exists t∗ y ≤ b such that F (y) F (t∗). Since F (t∗) ≥ F (t) F (bn), and F (bn) ≥ F (z) for all bn ≤ z ≤ b, we see that y cannot exceed bn, and thus lies in A, but this contradicts the fact that t∗ is the supremum of A. The case when an = a is similar and is left to the reader the only difference is that we can no longer assert that F (y) ≤ F (an) for all an ≤ y ≤ b, and so do not have the upper bound F (bn) ≤ F (an). 14Author’s note: I have deliberately omitted including such pictures in the text, as I feel that it is far more instructive and useful for the reader to directly create a personalised visual aid for these results.

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