Review of Lagrangian and Eulerian descriptions: Jacobian, Euler’s identity and Reynold’s transport theorem. The continuity equation and incompressibility condition.

这是一份Bath巴斯大学MA40256作业代写的成功案

Let $\tilde{Q}{j}$ denote a cube strictly contained in $Q{j}$ such that $\left|Q_{j}\right| \leq\left|\tilde{Q}{j}\right|+$ $\epsilon / 2^{j}$, where $\epsilon$ is arbitrary but fixed. Then, for every $N$, the cubes $\tilde{Q}{1}, \tilde{Q}{2}, \ldots, \tilde{Q}{N}$ are disjoint, hence at a finite distance from one another, and repeated applications of Observation 4 imply
$$
m_{}\left(\bigcup_{j=1}^{N} \tilde{Q}{j}\right)=\sum{j=1}^{N}\left|\tilde{Q}{j}\right| \geq \sum{j=1}^{N}\left(\left|Q_{j}\right|-\epsilon / 2^{j}\right) .
$$
Since $\bigcup_{j=1}^{N} \tilde{Q}{j} \subset E$, we conclude that for every integer $N$, $$ m{}(E) \geq \sum_{j=1}^{N}\left|Q_{j}\right|-\epsilon
$$
In the limit as $N$ tends to infinity we deduce $\sum_{j=1}^{\infty}\left|Q_{j}\right| \leq m_{}(E)+\epsilon$ for every $\epsilon>0$, hence $\sum_{j=1}^{\infty}\left|Q_{j}\right| \leq m_{}(E)$. Therefore, combined with Observation 2, our result proves that we have equality.

MA40256 COURSE NOTES ：

So, suppose $F$ is compact (so that in particular $m_{}(F)<\infty$ ), and let $\epsilon>0$. By Observation 3 we can select an open set $\mathcal{O}$ with $F \subset \mathcal{O}$ and $m_{}(\mathcal{O}) \leq m_{}(F)+\epsilon$. Since $F$ is closed, the difference $\mathcal{O}-F$ is open, and by Theorem $1.4$ we may write this difference as a countable union of almost disjoint cubes $$ \mathcal{O}-F=\bigcup_{j=1}^{\infty} Q_{j} $$ For a fixed $N$, the finite union $K=\bigcup_{j=1}^{N} Q_{j}$ is compact; therefore $d(K, F)>0$ (we isolate this little fact in a lemma below). Since $(K \cup$ $F) \subset \mathcal{O}$, Observations 1,4 , and 5 of the exterior measure imply $$ \begin{aligned} m_{}(\mathcal{O}) & \geq m_{}(F)+m_{}(K) \
&=m_{}(F)+\sum_{j=1}^{N} m_{}\left(Q_{j}\right)
\end{aligned}
$$