Prove that$ \mu(A) \geq \lambda^{+}(A) $and $\nu(A) \geq \lambda^{-}(A)$ for every $A \in A$.

Question

I don't know how, however.

Let $(X;A)$ be a measurable space. Suppose $\lambda= \mu - \nu$, where $\mu$ and $\nu$ are finite positive measures. Prove that $\mu(A) \geq \lambda^{+}(A)$ and $\nu(A) \geq \lambda^{-}(A)$ for every $A \in A$.

Answer 1

Let $P, N$ be the positive, respectively the negative set given from the Hahn decomposition theorem ($N\cap P=\emptyset$, $N\cup P=X$ and $\lambda^+(A) = \lambda(A\cap P)$, $\lambda^-(A) = \lambda(A\cap N)$).

We compute $$\mu(A) \geq \mu(A\cap P) = \lambda(A\cap P) + \nu(A\cap P) = \lambda^+(A) + \nu(A\cap P) \geq \lambda^+(A).$$ On the other hand, we have $$ \nu(A) \geq \nu(A\cap N) = \mu(A\cap N) - \lambda(A\cap N) = \mu(A\cap N) + \lambda^-(A) \geq \lambda^-(A). $$