# The extension of a premeasure on an algebra

I got stuck with a proof from the text by Stein and Shakarchi. Please help me out. Thank you so much. Please find the picture below and note the words marked with red.

Q1: Why $$\nu(F)\leq\sum_{j=1}^{\infty}\nu(E_j)$$? If $$\nu$$ were induced by the exterior measure, the inequality would readily hold. However, I don't see any evidence that leads to the assumption.

Q2: Why $$\nu(E-F)\leq\mu(E-F)$$? Here's my calculation.

\begin{align*} \mu(E-F)-\nu(E-F)&=\mu(E)-\mu(F)-( \nu(E)-\nu(F) )\\ &=\nu(F)-\mu(F)\\ &\leq 0 \end{align*}

Did you see that? I got an inequality that goes in the other direction! What's wrong with me? I feel very confused. Please give me a hand.

$$\mu$$ is a measure and $$F \subseteq \cup_j E_j$$. This implies $$\nu (F) \leq \nu (\cup_j E_j) \leq \sum_k \nu(E_j)$$.

In your second question why is $$\nu (F)\leq \mu (F)$$?

• Does any abstract measure possess monotonicity? We obtain $\nu (F)\leq \mu (F)$ once we finish Q1.
– Boar
Commented Feb 20, 2020 at 9:12
• Yes, surely. @Steve Commented Feb 20, 2020 at 9:14

A1:

If $$E_1\subseteq E_2$$, then $$\nu(E_2)=\nu(E_1)+\nu(E_2\backslash E_1)$$. Hence, we have $$\nu(E_1)\leq\nu(E_2)$$ even if $$\nu(E_1)=\infty$$. By using the monotonic property, we can now answer Q1. Write $$\cup E_j=E_1\cup(E_2\backslash E_1)\cup(E_3\backslash E_1\backslash E_2)\cup\cdots.$$ Then \begin{align*} \nu(F)&\leq\nu(\cup E_j)\\ &=\nu(E_1)+\nu(E_2\backslash E_1)+\nu(E_3\backslash E_1\backslash E_2)+\cdots\\ &\leq\nu(E_1)+\nu(E_2)+\nu(E_3)+\cdots\\ &=\sum_{j=1}^{\infty}\nu(E_j). \end{align*} A2:

To obtain the inequality in Q2, one can plug $$Q=E\backslash F$$ into $$\nu(Q)\leq\mu(Q)$$, which is confirmed at an early stage. Note that $$Q$$ doesn't need to have finite measure.