## Bounded Linear Functionals and sets of measure zero

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 $f,g \in L_1$.

-
 If $M$ has measure zero then $1_M = 0$ $\mu$-a.e.so $1_M = 0$ as elements in $L^1 (\mu)$. Not sure if this is a research-level question, though. – Mark Schwarzmann Jul 28 2011 at 10:43 I concluded the same since $\int 1_M d\mu = 0$ but was not sure what the zero element of $L^1$ would be. Thanks for the really quick response. That answers my question and makes sense. – Ramesh Kadambi Jul 28 2011 at 10:50 I apologize, I will post to math.stackexchange.com. I did not realize this was only research level questions. – Ramesh Kadambi Jul 29 2011 at 0:27