Bounded Linear Functionals and sets of measure zero - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:36:04Z http://mathoverflow.net/feeds/question/71476 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71476/bounded-linear-functionals-and-sets-of-measure-zero Bounded Linear Functionals and sets of measure zero Ramesh Kadambi 2011-07-28T10:38:01Z 2011-07-28T10:51:45Z <p>I am teaching myself measure theory from Bartles "The elements of integration and lebesgue measure". In order to prove Riesz Representation Theorem he defines a set function $\lambda(E) = G(1_E)$. Where $G$ is a linear bounded functional, and $1_E$ is the usual characteristic function. In order to show that $\lambda$ is absolutely continuous with $\mu$, He claims that $\lambda$ defined as before is zero on a set of measure zero. We are working under a measure space $(X, \sigma(X), \mu)$ and $G$ is a bounded linear functional on $L_1(X, \sigma(X), \mu)$. The question is why is $\lambda(M) = G(1_M)$ zero if $M$ is a set of measure zero? I am unable to figure out how this comes about from the fact I know about linear functionals $G(af + bg) = aG(f) + bG(g)$ where $f,g \in L_1$.</p>