format updates

content/Math4121/Math4121_L30.md

@@ -20,7 +20,8 @@ $$
 m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Inner regularity:
 
@@ -34,7 +35,7 @@ $$
 
 So $m_i(S)<m(K)+\epsilon$. Since $\epsilon$ is arbitrary, $m_i(S)\leq m_e(S)$.
 
-QED
+</details>
 
 We can approximate $m(S)$ from outside by open sets. If we are just concerned with "approximating" $m(S)$, we can use finite union of intervals.
 
@@ -58,7 +59,8 @@ $$
 
 where $U=\bigcup_{j =1}^n I_j$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon>0$ and $m(V)<m(S)+\frac{\epsilon}{2}$. Let $K\subseteq S$ be closed set such that $m(S)-\frac{\epsilon}{2}<m(K)$. $V$ is an open cover of closed and bounded set $K$. By Heine-Borel theorem, $K$ has a finite subcover. Let $I_1,I_2,\cdots,I_n$ be the open intervals in the subcover.
 
@@ -68,7 +70,7 @@ $$
 m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon
 $$
 
-QED
+</details>
 
 Recall $\{T_j\}_{j=1}^\infty$ are disjoint measurable sets. Then $T=\bigcup_{j=1}^\infty T_j$ is measurable and