Showing that a supremum is a maximum

[whocaresdr]

Homework Statement

Let f be a positive integrable continuous function on ℝ. Fix a measurable set E such that E [itex]\subset[/itex] [0,1]. Let

[itex] s = sup_{β \subset ℝ} [\int_{E} f_{β}(x)dx] [/itex]

where

[itex]f_{β}(x) = f(x + β). [/itex]

Show that s is actually a maximum (not just a supremum) that is, there is at least one β which gives the supremum.

Homework Equations

(I think they're included in the problem statement...)

The Attempt at a Solution

This was an old exam problem from when I took an intro analysis course (Royden) a couple of years ago (2010). I remember I missed something on this exam, but a solution was never posted and the professor never offered one in office hours when I asked. I came across this again and remained puzzled at what exactly had made my proof incorrect... so I'd welcome any thoughts. I know this isn't an attempt per se, but any guidance to start would be great.

 

[kurt.physics]

This is what I had written: Let E be a measurable subset of [0,1]. Let f be a continuous and integrable function. Let s = supβ[∫Efβdx] We note that fβ(x) = f(x + β). Let {βn} be any sequence such that ∫Efβndx → s as n → ∞. We must show that there exists at least one β which gives the supremum. We consider the set of all β for which ∫Efβdx ≥ s – 1/n. This set is non-empty since it contains βn. Next, this set is bounded above by some β*, since f is continuous on ℝ. Thus, the sequence {βn} is bounded above by β*. By the Bolzano-Weierstrass Theorem, there exists a subsequence {βkn} of {βn} which converges to some β**. Then, by continuity of f, we have ∫Efβ**dx → ∫Efβkn dx as k → ∞. But since {βkn} converges to β**, we have ∫Efβkn dx → ∫Efβ**dx as k → ∞. Thus, ∫Efβ**dx = limk→∞∫Efβkn dx = limn→∞∫Efβndx = s. Therefore, β** gives the supremum.