My question is that if $\sigma_{n}(x)\to f(x)$ almost everywhere, where $f(x)$ is essentially bounded function and $|\sigma_{n}(x)|\leq K$ then how can we prove that $|f(x)|\leq K$
Terminology between essentially bounded function and bounded function.
0
$\begingroup$
real-analysis
-
0Kns last inequality you mean almost surely ? – 2012-06-15
-
0yes i want to show that $|f(x)|\leq K$ a.e.. – 2012-06-15
-
0Just take the limit in the inequality $|\sigma_{n}(x)|\leq K$ – 2012-06-15
-
0So we get $|f(x)|\leq K$ a.e., it implies $f(x)$ is essentially bounded. does it mean $f(x)$ is bounded? – 2012-06-15
-
0No see the William's answer. It is not possible, in general, to show that $f$ is bounded. Another very simple example can be constructed by using a Dirac delta measure. – 2012-06-15