Subadditivity of Measure

2.1.3 - Real Analysis: Royden

Let

$$\{E_k\}_{k=1}^{\infty}$$

be a countable collection of sets in $\mathcal{A}$. Prove that

$$m\left(\bigcup_{k=1}^{\infty} E_k \right) \leq \sum_{k=1}^{\infty}m(E_k)$$

Proof:

We prove by induction. Consider the base case for the sets $E_1$ and $E_2$. We have

$$E_2 \cup E_1 = E_2 \dot \cup (E_1\setminus(E_2 \cap E_1))$$

and thus

$$ \begin{aligned} m(E_2 \cup E_1) &= m(E_2) + m(E_1\setminus(E_2 \cap E_1)) \\ &\leq m(E_2) + m(E_1) \end{aligned} $$

The second line is justified by yesterday’s proof of monotonicity. For the induction step, suppose that

$$m \left(\bigcup_{k=1}^{n} E_k \right) \leq \sum_{k=1}^{n}m(E_k)$$

For ease of notation, let

$$U_n = \bigcup_{k=1}^{n} E_k$$

Now, consider

$$ \begin{aligned} \bigcup_{k=1}^{n+1} E_k &= E_{n+1} \cup U_n \\ &= E_{n+1} \dot \cup (U_n\setminus(U_n \cap E_{n+1})) \end{aligned} $$

and thus

$$ \begin{aligned} m\left(\bigcup_{k=1}^{n+1} E_k\right) &\leq m(E_{n+1}) + m(U_n) \\ &\leq m(E_{n+1}) + \sum_{k=1}^{n}m(E_k) \\ &= \sum_{k=1}^{n+1}m(E_k) \end{aligned} $$

Hence, we have that

$$m\left(\bigcup_{k=1}^{\infty} E_k \right) \leq \sum_{k=1}^{\infty}m(E_k)$$