I am reading an introduction to measure theory, which starts by defining $\sigma$-rings then additive set functions and their properties which are given without proof. I was able to prove two of them which are very easy $\phi(\emptyset)=0$ and $\phi(A_1\cup \ldots\cup A_n)=\displaystyle\sum_{i=1}^nA_i$.

There are three more which I couldn't do which are :

1- $\phi(A_1\cup A_2)+\phi(A_1 \cap A_2)=\phi(A_1)+\phi(A_2)$ for any two sets $A_1 $ and $A_2$.

2- If $\phi(A) \ge 0$ for all $A$ and $A_1 \subset A_2$ then $\phi(A_1) \le \phi(A_2)$.

3- If $A_1 \subset A_2$ and $|\phi(A_1)|< + \infty$ then $\phi(A_2-A_1)=\phi(A_2)-\phi(A_1)$.

Where $\phi$ is an additive set function.

I need some hints on how to proceed.

  • $\begingroup$ For 1 and 2, write $A_2=A_1\sqcup(A_2\setminus A_1)$. For 3, write $A\cup B-(A\setminus A\cap B)\sqcup (B\setminus A\cap B)\sqcup A\cap B$. Then apply $\phi$ observing the unions are disjoint. $\endgroup$
    – Julien
    Apr 9, 2013 at 20:51
  • $\begingroup$ @julien I was able to solve them and wrote the answer below. Are they correct? $\endgroup$
    – user10444
    Apr 9, 2013 at 20:52
  • 2
    $\begingroup$ Hmmm.... $A_2\ne A_1\cup(A_2\backslash A_1)$ $\endgroup$
    – David
    May 8, 2013 at 4:54

2 Answers 2


I was able to use the third proposition to prove the first.

Let $A \subset B$. We have $(A \setminus B) \cap B = \emptyset$ then since $\phi$ is additive we have $\phi(A \setminus B) + \phi(B)=\phi((A-B) \cup B)=\phi(A) $ . Then $\phi(A \setminus B)=\phi(A)-\phi(B)$, this proves the second one.

Now for the first : $A \cup B= A\setminus (A\cap B) \cup B \setminus (A\cap B) \cup(A \cap B)$. Then $\phi(A \cup B)=\phi(A\setminus (A\cap B) \cup B \setminus (A\cap B) )= \phi(A)-\phi(A\cap B)+\phi(B)- \phi(A\cap B) + \phi(A \cap B)$

Hence upon adding $\phi(A \cap B) $ to both sides we get $\phi(A \cup B) + \phi(A \cap B)= \phi(A) + \phi (B)$. Can anyone help with the second one?

Edit: Here's the second one. Since $\phi(A) > 0$ for all sets then $\phi(A\setminus B)>0$ hence by the second property $\phi(A)-\phi(B) > 0$ when $B \subset A$.

  • $\begingroup$ Your union formula is not correct. You miss a $A\cap B$. $\endgroup$
    – Julien
    Apr 9, 2013 at 20:54
  • $\begingroup$ @julien Thank you, I edited it in. $\endgroup$
    – user10444
    Apr 9, 2013 at 21:01

I'm reading the same book I guess :). This is my try:

The core point is that $\phi (A_1 \cup A_2) = \phi [A_1 \cup (A_2 - A_1)]$ implies $\phi (A_2 - A_1) = \phi (A_1 \cup A_2) - \phi (A_1)$, by the properties you stated.

3 follows immediately, since $A_1 \subset A_2$ implies $\phi (A_1 \cup A_2) = \phi(A_2)$.

2 follows from $ 0 \leq \phi(A_2 - A_1) = \phi (A_1 \cup A_2) - \phi (A_1) = \phi (A_2) - \phi (A_1) $.

1 is a little more involved: $ \phi(A_1 \cup A_2) + \phi(A_1 \cap A_2) = \phi[(A_1 - A_2) \cup A_2] + \phi[A_1 - (A_1 - A_2)] = \phi(A_1 - A_2) + \phi (A_2) + \phi[A_1 \cup (A_1 - A_2)] - \phi(A_1 - A_2) = \phi (A_2) + \phi[A_1 \cup (A_1 - A_2)] = \phi (A_2) + \phi(A_1) $.

  • $\begingroup$ Do you know where the $|\phi(A_{1})| < +\infty$ part comes into play? $\endgroup$
    – user695931
    Aug 18, 2019 at 12:23

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