Update documentation

_sources/book/mira/000-exercises.md

@@ -1510,7 +1510,7 @@ $$X_1 \times Y_1, X_1 \times Y_2, X_2 \times Y_1, \dots.$$
 
 This sequence is disjoint and its union equals $S \times T.$
 Also, $\omega(R_k) = (\mu \times \nu)(R_k) < \infty$ for all $k \in \mathbb{R}.$
-Therefore, defining $E = \cap_{n = 1}^\infty,$ we have
+Therefore, defining $E = \cap_{n = 1}^\infty E_n,$ we have
 
 $$\begin{align}
 \omega(E) &= \sum_{j = 1}^\infty \omega(R_j \cap E) \\
@@ -1520,7 +1520,8 @@ $$\begin{align}
           &= \lim_{n \to \infty} \omega(E_n),
 \end{align}$$
 
-and, by a similar argument, we have
+where in the third line we have used the bounded convergence theorem.
+By a similar argument, we have
 
 $$(\mu \times \nu)(E) = \lim_{n \to \infty} (\mu \times \nu)(E_n).$$
 

book/mira/000-exercises.html

@@ -1782,7 +1782,7 @@ <h2>Chapter 5.A<a class="headerlink" href="#chapter-5-a" title="Link to this hea
 \[X_1 \times Y_1, X_1 \times Y_2, X_2 \times Y_1, \dots.\]</div>
 <p class="sd-card-text">This sequence is disjoint and its union equals <span class="math notranslate nohighlight">\(S \times T.\)</span>
 Also, <span class="math notranslate nohighlight">\(\omega(R_k) = (\mu \times \nu)(R_k) &lt; \infty\)</span> for all <span class="math notranslate nohighlight">\(k \in \mathbb{R}.\)</span>
-Therefore, defining <span class="math notranslate nohighlight">\(E = \cap_{n = 1}^\infty,\)</span> we have</p>
+Therefore, defining <span class="math notranslate nohighlight">\(E = \cap_{n = 1}^\infty E_n,\)</span> we have</p>
 <div class="math notranslate nohighlight">
 \[\begin{split}\begin{align}
 \omega(E) &amp;= \sum_{j = 1}^\infty \omega(R_j \cap E) \\
@@ -1791,7 +1791,8 @@ <h2>Chapter 5.A<a class="headerlink" href="#chapter-5-a" title="Link to this hea
           &amp;= \lim_{n \to \infty} \sum_{j = 1}^\infty \omega(R_j \cap E_n) \\
           &amp;= \lim_{n \to \infty} \omega(E_n),
 \end{align}\end{split}\]</div>
-<p class="sd-card-text">and, by a similar argument, we have</p>
+<p class="sd-card-text">where in the third line we have used the bounded convergence theorem.
+By a similar argument, we have</p>
 <div class="math notranslate nohighlight">
 \[(\mu \times \nu)(E) = \lim_{n \to \infty} (\mu \times \nu)(E_n).\]</div>
 <p class="sd-card-text">Putting these results together, we have <span class="math notranslate nohighlight">\(\omega(E) = (\mu \times \nu)(E),\)</span> so <span class="math notranslate nohighlight">\(\mathcal{M}\)</span> is a monotone class.