Cfrgtkky

(a) For any bounded Baire function, $f$, on a compact Hausdorff space $X$, prove that $||f||_{infty}:=infleft{sup_x |g(x)|: f-g=0text{ for a.e. } xright}$ exists, defines a seminorm, and equals $sup_{x} |f(x)|$ if $fin C(X)$. Prove that

$||f||_{infty}=supleft{a:mu(left{x:|f(x)|>aright})>0right}$

(b) For any bounded Baire function $f$ and any $mu_{l}$ for normalized positive functional $l$, we have that

$|l(f)|leq ||f||_{infty}$

(c) If $f_n,f$ are Bounded Baire functions and $||f_n-f||_{infty}to 0$, prov that $l(f)to l(f_n)$.

For b) I have: $|f|leq |f|_{infty}$ then $|l(f)|leq l(|f|)leq ||f||_{infty}l(1)=||f||_{infty}$ because $l$ is normalized.

For c) I have, Using (b), $|l(f-f_n)|leq ||f-f_n||_{infty}to 0 $

How prove $||f||_{infty}:=infleft{sup_x |g(x)|: f-g=0text{ for a.e. } xright}$ exists? Some hint?

It is obvious that $f=g$ a.e. implies $|f(y)| leq sup_x|g(x)|$ a.e. so $|f(y)| leq RHS$ a.e. Hence LHS $leq $RHS. On the other hand if $g(x)=f(x)$ when $|f(x)| leq |f|_{infty}$ and $0$ otherwise then $f=g$ a.e. and $sup_x|g(x)| leq |f|_{infty}$ so RHS $leq$ LHS.