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.


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.


1 Answer 1


Your proof is definitely correct. I'm not sure what you mean by "standard," but I think you could safely say that it is – it seems like the most straightforward approach, and also, the typical proof that I remember for subadditivity in general goes like this (see Folland's, "Real analysis: Modern techniques and their applications", Theorem 1.8b):

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 $E_2 = E$.

  • $\begingroup$ When $n = \infty$, how can I define $A_j$? and how does the index j move? $\endgroup$
    – Curious
    Apr 27, 2021 at 5:09

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