- #1

- 17

- 0

Prove that [tex]\mu(\left\{f>h\right\})\leq A[/tex].

My Work:

Suppose not, then [tex]\mu(\left\{f>h\right\}) > A[/tex].

From the triangle inequality for measures we get

[tex]\mu(\left\{f>h\right\}) = \mu(\left\{f-f_{n}+f_{n}>h\right\})

\leq\mu(\left\{f-f_{n}>0\right\}) + \mu(\left\{f_{n}>h\right\}) [/tex].

So [tex] A<\mu(\left\{f-f_{n}>0\right\}) + \mu(\left\{f_{n}>h\right\})

\leq\mu(\left\{f-f_{n}>0\right\}) + A [/tex].

Taking limits on both sides (n->00)yields:

[tex] A < 0 + A \Rightarrow\Leftarrow[/tex].

I do not have much of a background for analysis, so any suggestions are welcome.