Let $\mathfrak{M}$ be a $\sigma$ algebra of subsets of $X$ and let $\mu:\mathfrak{M}\rightarrow\mathbb{R}$ be a countably additive measure.$A \in \mathfrak{M}, \mu(A) = \mu(X), f \in L_1(A)$. Prove that $ f \in L_1(X)$ and $\int\limits_{X}{fd\mu} = \int\limits_{A}{fd\mu}$
Since $f$ is integrable on $A$ then it is measurable on this set, hence it is also measurable on $X$. Is not this enough to claim that it is also integrable on $X$?
How do I prove the equality? I do not need a complete solution. A hint is enough. Thank you
You have to assume that $f$ is measurable on $X$. $\int_X f\, d\mu=\int_A f \,d\mu+\int_{X\setminus A} f \,d\mu$ and the second term is $0$. [Any measurable function is integrable over a measurable set of measure $0$ and value of the integral is $0$].