Usbrth

For any $ f in M^+(mathbbR)$ (measurable positive functions), the function $$lambda: mathcalA rightarrow mathbbR\
mathcalA text is the set of measurable funtions$$

given by $ lambda(E) = int_Ef $ is a measure.

I already proved:

i) $lambda(emptyset) = int_emptysetf = 0$

ii) let $A subseteq B in mathcalA$ then $$f|_A le f|_B$$'
so $$ int f|_A le int f|_B $$ that is $$ lambda(A) = int_Af=int f|_A = int fX_A le int fX_B = int f|_B = int_B f = lambda(B)$$

Now I am stuck in sigma additivity

iii) let $A,B in mathcalA$ so $ lambda(Acup B) = lambda(A) +lambda(B) $ if the sets are disjoint then I think I can do it, but can we do that assumption? any help to prove this?

Use the dominated convergence theorem
https://en.wikipedia.org/wiki/Dominated_convergence_theorem

on

$f_N=fchi_A_N $ where $ A_N=bigcup_i=0^N E_i $

$int f_N = int fchi_A_N=int _bigcup_i=0^N E_if=sum_i=0^Nint_E_if$

check that it satisfies all the conditions when you take $g=f$

The theorem implies

$lim _nrightarrow inftyint f_N=lim _nrightarrow inftysum_i=0^Nint_E_if = sum_i=0^inftyint_E_if = int f$

In general if $(f_n)_n$ denotes a sequence of measurable functions that all take values in $[0,infty]$ then:

• Also $f:=sum_n=1^inftyf_n$ is a properly defined measurable function that takes values in $[0,infty]$.


• For every $n$ the integral $int f_n$ is properly defined.


• Also the integral $int f$ is properly defined and satisfies $int f=sum_n=1^inftyint f_n$.

This under the convention that for $a,bin[0,infty]$ we have $a+b:=infty$ if $inftyina,b$.

Concerning the third bullet we have $fgeqsum_n=1^mf_n$ and consequently $int fgeqintsum_n=1^mf_n=sum_n=1^mint f_n$ for every $m$ so that $int fgeqsum_n=1^inftyint f_n$.

Conversely the lemma of Fatou can be used: $$int f=intliminf_mtoinftysum_n=1^mf_nleqliminf_mtoinftyintsum_n=1^mf_n=liminf_mtoinftysum_n=1^mint f_n=sum_n=1^inftyint f_n$$

Using this we find for measurable and disjoint sets $A_n$ that: $$lambdaleft(bigcup_n=1^inftyA_nright)=int fmathbf1_bigcup_n=1^inftyA_n=intsum_n=1^inftyfmathbf1_A_n=sum_n=1^inftyint fmathbf1_A_n=sum_n=1^inftylambda(A_n)$$