I'm reading Stein's Real Analysis, and I'm stuck on the proof of Theorem 3.4, Ch.1, Pg. 22. Here's the statement and proof.
$\textbf{Theorem 3.4.} \textit{ Suppose $E$ is a measurable subset of $\mathbb{R}^d$ such that $m(E)<\infty.$}$
$\textit{Then, for every $\varepsilon >0$, there exists a finite union $F=\bigcup_{j=1}^N Q_j$ of closed cubes such that }m(E\triangle F)\leq \varepsilon.$
Proof.
By definition of the infimum we may choose a family of closed cubes $\{Q_j\}_{j=1}^\infty$ that cover $E$, satisfying
\begin{equation*}
\sum_{j=1}^\infty |Q_j| \leq m(E) + \varepsilon/2.
\end{equation*}
Since $m(E)<\infty,$ the series $\sum |Q_j|$ converges, hence by the Cauchy criterion there exists $N>0$ such that $\sum_{j=N+1}^\infty |Q_j| < \varepsilon/2.$ Let $F=\bigcup_{j=1}^N Q_j.$ Then
\begin{align*}
m(E\triangle F) &= m(E-F) + m(F-E)\\
&\leq m\left(\bigcup_{j=N+1}^\infty |Q_j|\right) + m\left( \bigcup_{j=1}^\infty Q_j – E\right)\\
&\leq \sum_{j=N+1}^\infty |Q_j| + \sum_{j=1}^\infty |Q_j| – m(E)\\
&\leq \varepsilon,
\end{align*}
as desired.
Question. I don't understand how we can make the transition from the second to third line, namely
$$ m\left(\bigcup_{j=1}^\infty Q_j – E\right) \leq \sum_{j=1}^\infty |Q_j| – m(E). $$
By writing $\bigcup_{j=1}^\infty Q_j = \left(\bigcup_{j=1}^\infty Q_j – E\right) \cup E,$ I discovered that
\begin{align*}
m\left(\bigcup_{j=1}^\infty Q_j\right)\leq m\left(\bigcup_{j=1}^\infty Q_j – E\right) + m(E) \implies m\left(\bigcup_{j=1}^\infty Q_j – E\right) \geq m\left(\bigcup_{j=1}^\infty Q_j \right) – m(E),
\end{align*}
but this inequality is in the opposite direction from what I want. 🙁
Would appreciate all help.
Best Answer
Let $A = \cup_{j = 1}^\infty Q_j$. Then $A\supset E$, so $m(A\setminus E) + m(E) = m(A\setminus E) + m(A\cap E) = m(A)$. Thus $m(A\setminus E) = m(A) - m(E)$. Now $m(A)\le \sum_{j = 1}^\infty |Q_j|$, so $m(A\setminus E) \le \sum_{j =1}^\infty |Q_j| - m(E)$, as desired.