Page 318 - DMTH401_REAL ANALYSIS

Real Analysis

Notes

Notes The above statements imply f (a)  M for all a  [–¥, ¥]. The converse is not true.
–1
Proof: The first one is clearly equivalent to the fourth one since f ([a, ¥]) = E \ f ([–¥, a]).
–1
–1
Similarly, the second and the third statements are equivalent. It remains to show the first two
statements are equivalent, but this follows immediately from
¥ 1 æ 1 ö ¥ 1 æ 1 ö
f ([a, ¥]) =  f - ç (a - , ] ÷ and f ([a, ¥]) =  f - ç [a + , ] . ÷
–1
–1
¥
¥
=
=
n 1 è n ø n 1 è n ø
Proposition: Let E  M, f: E  [–¥, ¥] and g: E  [–¥, ¥]. If f = g almost everywhere on E then
the measurability of f and g are the same.
Proof: Simply note that
m*({x E : f(x) > a }  {x  E : g(x) > a})  m*({x  E : f(x)  g(x)}) = 0.
This implies the measurability of the sets {x  E: f(x) > a} and {x  E : g(x) > a} are the same.
Proposition: Let f, g be measurable extend real-valued functions on E  M. Then the following
functions are all measurable on E:

f + c, cf, f ± g, fg
where c  .

Notes One may worry that cf, f ± g, fg may not be defined at some points (for example, if
f = ¥and g = –¥ then f + g is meaningless). There are two ways to deal with this problem.

1. Adopt the convention 0· ¥ = 0.
2. Assume f, g are finite almost everywhere or cf, f ± g, fg are meaningful almost everywhere.

Proof: We only prove f + g and fg are measurable, since the others are easy or similar.
To prove f + g is measurable, one should consider the set
E = {x  E : f(x) + g(x) > a}
a
= {x  E : f(x)> a – g(x)}
=  {x E : f(x) > r > a – g(x)}

r
=  {x E : f(x) > r}  {x E : r > a – g(x)}


r
If f(x) = ¥ or g(x) = ¥ then x  E by convention. Now E  M because E is countable union of
a a a
measurable sets.
Next, we prove f is measurable. For a  0,
2
2
{x  E: f (x) > a} = {x  E : f(x) > a }  {x  E : f(x)< – a }
2
2
is measurable. For a < 0, {x  E : f (x) > a } = E is also measurable. Therefore, f is measurable and
it is valid even if f takes values ± ¥.

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