Page 319 - DMTH401_REAL ANALYSIS

Unit 26: Lebesgue Measure

So, if f and g are assumed to be finite, then Notes
1
2
2
fg = [(f + g) – f – g ]
2
2
is measurable on E.

Task Find two measurable functions f, g from  to  such that f o g is not measurable.
Proposition: Let { } n be measurable extended real-valued functions on a measurable set E.
f
n
Then
f  f …  f , sup f , lim f n
1 2 n n
n n¥
are all measurable on E. Similar results hold if , sup and lim are replaced by , inf, and lim.
Proof: Simply note that

¥
1
-
(f  f   f ) ((a, ¥)) =  f ((a, ))
–1
¥
1 2 n k
k 1
=
- 1
¥
æ ö ((a, )) =  - 1
¥
ç sup f n÷ ¥ f ((a, ))
è n ø k 1
k
=
æ ö
lim f = inf ç sup f k ÷ ø
n


n¥ N è k N
Theorem 7: Let E  M with m(E) < ¥, f: E  [–¥,¥] be measurable and finite almost everywhere.
For any  > 0, there is a simple function  such that
|f –|< on E except on a set of measure less than .
Notes If E = [a, b] is closed and bounded interval, we can find a step function g and a
continuous function h play the role of . This is because simple function can be approximated
by step function and step function can be approximated by continuous function.
If f satisfies an additional condition m  f  M, then , g, and h can be chosen to be bounded below
by m and above by M.
The condition m(E) < ¥ in Littlewood’s 2nd Principle is essential. You can see if this condition is
dropped then taking f(x) = x will give a counter example.
To prove Littlewood’s 2nd Principle, we introduce a lemma.
Lemma: Let { } n be a sequence of measurable subsets of  (or any measure space ) such that
3
F
n
F  F  .
1 2
Denote F =  n F . If m(F ) < ¥ then
¥ n 1
m(F ) = lim m(F ).
n
¥ n¥
Proof: Write F = F  (F \F )  (F \F )   as disjoint union and use the countable additivity of m.
1 ¥ 1 2 2 3

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