# How do we prove the finite subadditivity property of the elementary measure?

So this is my question: how do we prove the finite subadditivity property of elementary measure based on the monotonicity and the finite additivity properties? Precisely speaking, I am interested in the following property: \begin{align*} m(E\cup F) \leq m(E) + m(F) \end{align*} whenever $$E$$ and $$F$$ are elementary sets.

MY ATTEMPT

Since $$E\cup F = (E\backslash F)\cup F$$ and $$E\backslash F\subseteq E$$, one has that \begin{align*} m(E\cup F) = m(E\backslash F) + m(F) \leq m(E) + m(F) \end{align*}

I am new to this. Is this approach standard? Any contribution is appreciated.

For a countable collection $$\{E_j\}_{j=1}^n$$ of measurable sets (possibly with $$n=\infty$$), define $$A_j := E_j \setminus \big(\bigcup_{k=1}^{j-1} E_k\big)$$ for all $$1 \leq j \leq n$$. Then the sets $$\{A_j\}_{j=1}^n$$ are measurable and mutually disjoint, and $$\bigcup_{j=1}^n A_j = \bigcup_{j=1}^nE_j$$, so $$m\big( \bigcup_{j=1}^n E_j \big) = m\big( \bigcup_{j=1}^n A_j \big) = \sum_{j=1}^n m(A_j) \leq \sum_{j=1}^n m(E_j).$$
This reduces exactly to the argument you gave in the case that $$n=2$$, $$E_1 = F$$, and $$E_2 = E$$.
• When $n = \infty$, how can I define $A_j$? and how does the index j move? Apr 27, 2021 at 5:09