• If \(m_*(E) = \infty\), then take \(B = {\mathbb{R}}^n\) since \(m({\mathbb{R}}^n) = \infty\).

• Suppose \(N \coloneqq m_*(E) < \infty\).

• Since \(m_*(E)\) is an infimum, by definition, for every \({\varepsilon}> 0\) there exists a covering by closed cubes \(\left\{{Q_i({\varepsilon})}\right\}_{i=1}^\infty \rightrightarrows E\) depending on \({\varepsilon}\) such that \begin{align*} \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon})} \right\rvert} < N + {\varepsilon} .\end{align*}

• For each fixed \(n\), set \({\varepsilon}_n = {1\over n}\) to produce such a covering \(\left\{{Q_i({\varepsilon}_n)}\right\}_{i=1}^\infty\) and set \(B_n \coloneqq\displaystyle\bigcup_{i=1}^\infty Q_i({\varepsilon}_n)\).

• The outer measure of cubes is equal to the sum of their volumes, so \begin{align*} m_*(B_n) = \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon}_n)} \right\rvert} < N + {\varepsilon}_n = N + {1\over n} .\end{align*}

• Now set \(B \coloneqq\displaystyle\bigcap_{n=1}^\infty B_n\).

  • Since \(E\subseteq B_n\) for every \(n\), \(E\subseteq B\)
  • Since \(B\) is a countable intersection of countable unions of closed sets, \(B\) is Borel.
  • Since \(B_n \subseteq B\) for every \(n\), we can apply subadditivity to obtain the inequality \begin{align*} E \subseteq B \subseteq B_n \implies N \leq m_*(B) \leq m_*(B_n) < N + {1\over n} {\quad \operatorname{for all} \quad} n\in {\mathbb{Z}}^{\geq 1} .\end{align*}

• This forces \(m_*(E) = m_*(B)\).

Suppose \(m_*(E) < \infty\).

• By (a), find a Borel set \(B\supseteq E\) such that \(m_*(B) = m_*(E)\)
• Note that \(E\subseteq B \implies B\displaystyle\bigcap E = E\) and \(B\displaystyle\bigcap E^c = B\setminus E\).
• By assumption, \begin{align*} m_*(B) &= m_*(B\displaystyle\bigcap E) + m_*(B\displaystyle\bigcap E^c) \\ m_*(E) &= m_*(E) + m_*(B\setminus E) \\ m_*(E) - m_*(E) &= m_*(B\setminus E) \qquad\qquad\text{since } m_*(E) < \infty \\ \implies m_*(B\setminus E) &= 0 .\end{align*}
• So take \(N = B\setminus E\); this shows \(m_*(N) = 0\) and \(E = B\setminus (B\setminus E) = B\setminus N\).

If \(m_*(E) = \infty\):

• Apply result to \(E_R\coloneqq E \displaystyle\bigcap[R, R+1)^n \subset {\mathbb{R}}^n\) for \(R\in {\mathbb{Z}}\), so \(E = {\textstyle\coprod}_R E_R\)
• Obtain \(B_R, N_R\) such that \(E_R = B_R \setminus N_R\), \(m_*(E_R) = m_*(B_R)\), and \(m_*(N_R) = 0\).
• Note that
  • \(B\coloneqq\displaystyle\bigcup_R B_R\) is a union of Borel sets and thus still Borel
  • \(E = \displaystyle\bigcup_R E_R\)
  • \(N\coloneqq B\setminus E\)
  • \(N' \coloneqq\displaystyle\bigcup_R N_R\) is a union of null sets and thus still null
• Since \(E_R \subset B_R\) for every \(R\), we have \(E\subset B\)
• We can compute \begin{align*} N = B\setminus E = \qty{ \displaystyle\bigcup_R B_R } \setminus \qty{\displaystyle\bigcup_R E_R } \subseteq \displaystyle\bigcup_R \qty{B_R\setminus E_R} = \displaystyle\bigcup_R N_R \coloneqq N' \end{align*} where \(m_*(N') = 0\) since \(N'\) is null, and thus subadditivity forces \(m_*(N) = 0\).

• Since \(E\) is measurable, \(E^c\) is measurable.
• Since \(E^c\) is measurable exists an open \(O\supseteq E^c\) such that \(m(O\setminus E^c) = 0\).
• Set \(B \coloneqq O^c\), then \(O\supseteq E^c \iff {\mathcal{O}}^c \subseteq E \iff B\subseteq E\).
• Computing measures yields \begin{align*} E\setminus B \coloneqq E\setminus {\mathcal{O}}^c \coloneqq E\displaystyle\bigcap({\mathcal{O}}^c)^c = E\displaystyle\bigcap{\mathcal{O}}= {\mathcal{O}}\displaystyle\bigcap(E^c)^c \coloneqq{\mathcal{O}}\setminus E^c ,\end{align*} thus \(m(E\setminus B) = m({\mathcal{O}}\setminus E^c) = 0\).
• Since \({\mathcal{O}}\) is open, \(B\) is closed and thus Borel.

d.irect Proof (Todo)

\todo[inline]{Try to construct the set.}

• Strategy: Borel-Cantelli.

• We’ll show that \(m(E) \displaystyle\bigcap[n, n+1] = 0\) for all \(n\in {\mathbb{Z}}\); then the result follows from \begin{align*} m(E) = m \qty{\displaystyle\bigcup_{n\in {\mathbb{Z}}} E \displaystyle\bigcap[n, n+1]} \leq \sum_{n=1}^\infty m(E \displaystyle\bigcap[n, n+1]) = 0 .\end{align*}

• By translation invariance of measure, it suffices to show \(m(E \displaystyle\bigcap[0, 1]) = 0\).

  • So WLOG, replace \(E\) with \(E\displaystyle\bigcap[0, 1]\).

• Define \begin{align*} E_j \coloneqq\left\{{x\in [0, 1] {~\mathrel{\Big\vert}~}\ \exists p\in {\mathbb{Z}}^{\geq 0} \text{ s.t. } {\left\lvert {x - \frac{p}{j} } \right\rvert} < \frac 1 {j^3}}\right\} .\end{align*}

  • Note that \(E_j \subseteq {\textstyle\coprod}_{p\in {\mathbb{Z}}^{\geq 0}} B_{j^{-3}}\qty{p\over j}\), i.e. a union over integers \(p\) of intervals of radius \(1/j^3\) around the points \(p/j\). Since \(1/j^3 < 1/j\), this union is in fact disjoint.

• Importantly, note that \begin{align*} \limsup_{j\to\infty} E_j \coloneqq\displaystyle\bigcap_{n=1}^\infty \displaystyle\bigcup_{j=n}^\infty E_j = E \end{align*}

  since

  \begin{align*} x \in \limsup_j E_j &\iff x \in E_j \text{ for infinitely many } j \\ &\iff \text{ there are infinitely many $j$ for which there exist a $p$ such that } {\left\lvert {x - {p\over j}} \right\rvert} < j^{-3} \\ &\iff \text{ there are infinitely many such pairs $p, j$} \\ &\iff x\in E .\end{align*}

• Intersecting with \([0, 1]\), we can write \(E_j\) as a union of intervals: \begin{align*} E_j =& \qty{0, {j^{-3}}} \quad {\textstyle\coprod}\quad B_{j^{-3}}\qty{1\over j} {\textstyle\coprod} B_{j^{-3}}\qty{2\over j} {\textstyle\coprod} \cdots {\textstyle\coprod} B_{j^{-3}}\qty{j-1\over j} \quad {\textstyle\coprod}\quad (1 - {j^{-3}}, 1) ,\end{align*} where we’ve separated out the “boundary” terms to emphasize that they are balls about \(0\) and \(1\) intersected with \([0, 1]\).

• Since \(E_j\) is a union of open sets, it is Borel and thus Lebesgue measurable.

• Computing the measure of \(E_j\):

  • For a fixed \(j\), there are exactly \(j+1\) possible choices for a numerator (\(0, 1, \cdots, j\)), thus there are exactly \(j+1\) sets appearing in the above decomposition.

  • The first and last intervals are length \(1 \over j^3\)

  • The remaining \((j+1)-2 = j-1\) intervals are twice this length, \(2 \over j^3\)

  • Thus \begin{align*} m(E_j) = 2 \qty{1 \over j^3} + (j-1) \qty{2 \over j^3} = {2 \over j^2} \end{align*}

• Note that \begin{align*} \sum_{j\in {\mathbb{N}}} m(E_j) = 2\sum_{j\in {\mathbb{N}}} \frac 1 {j^2} < \infty ,\end{align*} which converges by the \(p{\hbox{-}}\)test for sums.

• But then \begin{align*} m(E) &= m(\limsup_j E_j) \\ &= m(\displaystyle\bigcap_{n\in {\mathbb{N}}} \displaystyle\bigcup_{j\geq n} E_j) \\ &\leq m(\displaystyle\bigcup_{j\geq N} E_j) \quad\text{for every } N \\ &\leq \sum_{j\geq N} m(E_j) \\ &\overset{N\to\infty}\to 0 \quad\text{} .\end{align*}

• Thus \(E\) is measurable as a subset of a null set and \(m(E) = 0\).

It suffices to consider the bounded case, i.e. \(E \subseteq B_M(0)\) for some \(M\). Then write \(E_n = B_n(0) \displaystyle\bigcap E\) and apply the theorem to \(E_n\), and by subadditivity, \(m^*(E) = m^*(\displaystyle\bigcup_n E_n) \leq \sum_n m^*(E_n) = 0\).

Lemma: \(f(x) = x^2, f^{-1}(x) = \sqrt{x}\) are Lipschitz on any compact subset of \([0, \infty)\).

Proof: Let \(g = f\) or \(f^{-1}\). Then \(g\in C^1([0, M])\) for any \(M\), so \(g\) is differentiable and \(g'\) is continuous. Since \(g'\) is continuous on a compact interval, it is bounded, so \({\left\lvert {g'(x)} \right\rvert} \leq L\) for all \(x\). Applying the MVT, \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = f'(c) {\left\lvert {x-y} \right\rvert} \leq L {\left\lvert {x-y} \right\rvert} .\end{align*}

Lemma: If \(g\) is Lipschitz on \({\mathbb{R}}^n\), then \(m(E) = 0 \implies m(g(E)) = 0\).

Proof: If \(g\) is Lipschitz, then \begin{align*} g(B_r(x)) \subseteq B_{Lr}(x) ,\end{align*} which is a dilated ball/cube, and so \begin{align*} m^*(B_{Lr}(x)) \leq L^n \cdot m^*(B_{r}(x)) .\end{align*}

Now choose \(\left\{{Q_j}\right\} \rightrightarrows E\); then \(\left\{{g(Q_j)}\right\} \rightrightarrows g(E)\).

By the above observation, \begin{align*} {\left\lvert {g(Q_j)} \right\rvert} \leq L^n {\left\lvert {Q_j} \right\rvert} ,\end{align*}

and so \begin{align*} m^*(g(E)) \leq \sum_j {\left\lvert {g(Q_j)} \right\rvert} \leq \sum_j L^n {\left\lvert {Q_j} \right\rvert} = L^n \sum_j {\left\lvert {Q_j} \right\rvert} \to 0 .\end{align*}

Now just take \(g(x) = x^2\) for one direction, and \(g(x) = f^{-1}(x) = \sqrt{x}\) for the other.

Lemma: \(E\) is measurable iff \(E = K {\textstyle\coprod}N\) for some \(K\) compact, \(N\) null.

Write \(E = K {\textstyle\coprod}N\) where \(K\) is compact and \(N\) is null.

Then \(\phi^{-1}(E) = \phi^{-1}(K {\textstyle\coprod}N) = \phi^{-1}(K) {\textstyle\coprod}\phi^{-1}(N)\).

Since \(\phi^{-1}(N)\) is null by part (a) and \(\phi^{-1}(K)\) is the preimage of a compact set under a continuous map and thus compact, \(\phi^{-1}(E) = K' {\textstyle\coprod}N'\) where \(K'\) is compact and \(N'\) is null, so \(\phi^{-1}(E)\) is measurable.

So \(\phi\) is a measurable function, and thus yields a well-defined map \(\mathcal L({\mathbb{R}}) \to \mathcal L({\mathbb{R}})\) since it preserves measurable sets. Restricting to \([0, \infty)\), \(f\) is bijection, and thus so is \(\phi\).

Claim: \(K\) is compact.

• It suffices to show that \(K^c \coloneqq[0, 1]\setminus K\) is open; Then \(K\) will be a closed and bounded subset of \({\mathbb{R}}\) and thus compact by Heine-Borel.

• Strategy: write \(K^c\) as the union of open balls (since these form a basis for the Euclidean topology on \({\mathbb{R}}\)).

  • Do this by showing every point \(x\in K^c\) is an interior point, i.e. \(x\) admits a neighborhood \(N_x\) such that \(N_x \subseteq K^c\).

• Identify \(K^c\) as the set of real numbers in \([0, 1]\) whose decimal expansion does contain a 4.

  • We will show that there exists a neighborhood small enough such that all points in it contain a \(4\) in their decimal expansions.

• Let \(x\in K^c\), suppose a 4 occurs as the \(k\)th digit, and write \begin{align*} x = 0.d_1 d_2 \cdots d_{k-1}~ 4 ~d_{k+1}\cdots = \qty{\sum_{j=1}^k d_j 10^{-j}} + \qty{4\cdot 10^{-k}} + \qty{\sum_{j=k+1}^\infty d_j 10^{-j}} .\end{align*}

• Set \(r_x < 10^{-k}\) and let \(y \in [0, 1] \displaystyle\bigcap B_{r_x}(x)\) be arbitrary and write \begin{align*} y = \sum_{j=1}^\infty c_j 10^{-j} .\end{align*}

• Thus \({\left\lvert {x-y} \right\rvert} < r_x < 10^{-k}\), and the first \(k\) digits of \(x\) and \(y\) must agree:

  • We first compute the difference: \begin{align*} x - y &= \sum_{i=1}^\infty d_j 10^{-j} - \sum_{i=1}^\infty c_j 10^{-j} = \sum_{i=1}^\infty \qty{d_j - c_j} 10^{-j} \\ \end{align*}
  • Thus (claim) \begin{align*} {\left\lvert {x-y} \right\rvert} &\leq \sum_{j=1}^\infty {\left\lvert {d_j - c_j} \right\rvert} 10^j < 10^{-k} \iff {\left\lvert {d_j - c_j} \right\rvert} = 0 \quad \forall j\leq k .\end{align*}
  • Otherwise we can note that any term \({\left\lvert {d_j - c_j} \right\rvert}\geq 1\) and there is a contribution to \({\left\lvert {x-y} \right\rvert}\) of at least \(1\cdot 10^{-j}\) for some \(j < k\), whereas \begin{align*} j < k \iff 10^{-j} > 10^{-k} ,\end{align*} a contradiction.

• This means that for all \(j \leq k\) we have \(d_j = c_j\), and in particular \(d_k = 4 = c_k\), so \(y\) has a 4 in its decimal expansion.

• But then \(K^c = \displaystyle\bigcup_x B_{r_x}(x)\) is a union of open sets and thus open.

Claim: \(K\) is nowhere dense and \(m(K) = 0\):

• Strategy: Show \(\qty{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}^\circ = \emptyset\).

• Since \(K\) is closed, \(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu = K\), so it suffices to show that \(K\) does not properly contain any interval.

• It suffices to show \(m(K^c) = 1\), since this implies \(m(K) = 0\) and since any interval has strictly positive measure, this will mean \(K\) can not contain an interval.

• As in the construction of the Cantor set, let

  • \(K_0\) denote \([0, 1]\) with 1 interval \(\left({4 \over 10}, {5 \over 10} \right)\) of length \(1 \over 10\) deleted, so \begin{align*}m(K_0^c) = {1\over 10}.\end{align*}
  • \(K_1\) denote \(K_0\) with 9 intervals \(\left({1 \over 100}, {5\over 100}\right), ~\left({14 \over 100}, {15 \over 100}\right), \cdots \left({94\over 100}, {95 \over 100}\right)\) of length \({1 \over 100}\) deleted, so \begin{align*}m(K_1^c) = {1\over 10} + {9 \over 100}.\end{align*}
  • \(K_n\) denote \(K_{n-1}\) with \(9^{n}\) such intervals of length \(1 \over 10^{n+1}\) deleted, so \begin{align*}m(K_n^c) = {1\over 10} + {9 \over 100} + \cdots + {9^{n} \over 10^{n+1}}.\end{align*}

• Then compute \begin{align*} m(K^c) = \sum_{j=0}^\infty {9^n \over 10^{n+1} } = {1\over 10} \sum_{j=0}^\infty \qty{9\over 10}^n = {1 \over 10} \qty{ {1 \over 1 - {9 \over 10 } } } = 1. \end{align*}

Claim: \(K\) has no isolated points:

• A point \(x\in K\) is isolated iff there there is an open ball \(B_r(x)\) containing \(x\) such that \(B_r(x) \subsetneq K^c\).

  • So every point in this ball should have a 4 in its decimal expansion.

• Strategy: show that if \(x\in K\), every neighborhood of \(x\) intersects \(K\).

• Note that \(m(K_n) = \left( \frac 9 {10} \right)^n \overset{n\to\infty}\to 0\)

• Also note that we deleted open intervals, and the endpoints of these intervals are never deleted.

  • Thus endpoints of deleted intervals are elements of \(K\).

• Fix \(x\). Then for every \(\varepsilon\), by the Archimedean property of \({\mathbb{R}}\), choose \(n\) such that \(\left( \frac 9 {10} \right)^n < \varepsilon\).

• Then there is an endpoint \(x_n\) of some deleted interval \(I_n\) satisfying \begin{align*}{\left\lvert {x - x_n} \right\rvert} \leq \left( \frac 9 {10} \right)^n < {\varepsilon}.\end{align*}

• So every ball containing \(x\) contains some endpoint of a removed interval, and thus an element of \(K\).

• Strategy: use approximation by simple functions to show absolute continuity and apply Radon-Nikodym

• Claim: \(\lambda \ll \mu\), i.e. \(\mu(E) = 0 \implies \lambda(E) = 0\).

  • Note that if this holds, by Radon-Nikodym, \(f = {\frac{\partial \lambda}{\partial \mu}\,} \implies d\lambda = f d\mu\), which would yield \begin{align*} \int g ~d\lambda = \int g f ~d\mu .\end{align*}

• So let \(E\) be measurable and suppose \(\mu(E) = 0\).

• Then \begin{align*} \lambda(E) \coloneqq\int_E f ~d\mu &= \lim_{n\to\infty} \left\{{\int_E s_n \,d\mu {~\mathrel{\Big\vert}~}s_n \coloneqq\sum_{j=1}^\infty c_j \mu(E_j),\, s_n \nearrow f}\right\} \end{align*} where we take a sequence of simple functions increasing to \(f\).

• But since each \(E_j \subseteq E\), we must have \(\mu(E_j) = 0\) for any such \(E_j\), so every such \(s_n\) must be zero and thus \(\lambda(E) = 0\).

\todo[inline]{What is the final step in this approximation?}

• Set \(g(x) = x^2\), note that \(g\) is positive and measurable.

• By part (a), there exists a positive \(f\) such that for any \(E\subseteq {\mathbb{R}}\), \begin{align*} \int_E g ~dm = \int_E gf ~d\mu \end{align*}

  • The LHS is zero by assumption and thus so is the RHS.

  • \(m \ll \mu\) by construction.

  • Note that \(gf\) is positive.

• Define \(A_k = \left\{{x\in X {~\mathrel{\Big\vert}~}gf \cdot \chi_E > {1 \over k} }\right\}\), for \(k\in {\mathbb{Z}}^{\geq 0}\)

• Then by Chebyshev with \(p=1\), for every \(k\) we have

\begin{align*} \mu(A_k) \leq k \int_E gf ~d\mu = 0 \end{align*}

• Then noting that \(A_k \searrow A \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}gf\cdot \chi_E(x) > 0}\right\}\), we have \(\mu(A) = 0\).

• Since \(gf\) is positive, we have \begin{align*} x\in E \iff gf\chi_E(x) > 0 \iff x\in A \end{align*} so \(E = A\) and \(\mu(E) = \mu(A)\).

• But \(m \ll \mu\) and \(\mu(E) = 0\), so we can conclude that \(m(E) = 0\).

• Claim: \(G\in {\mathcal{M}}\).

  • Claim: \begin{align*} G = \qty{ \displaystyle\bigcap_{N=1}^\infty \displaystyle\bigcup_{n=N}^\infty E_n}^c = \displaystyle\bigcup_{N=1}^\infty \displaystyle\bigcap_{n=N}^\infty E_n^c .\end{align*}

    • This follows because \(x\) is in the RHS \(\iff\) \(x\in E_n^c\) for all but finitely many \(n\) \(\iff\) \(x\in E_n\) for at most finitely many \(n\).

  • But \({\mathcal{M}}\) is a \(\sigma{\hbox{-}}\)algebra, and this shows \(G\) is obtained by countable unions/intersections/complements of measurable sets, so \(G\in {\mathcal{M}}\).

• Claim: \(\mu(G) = 0\).

  • We have \begin{align*} \mu(G) &= \mu\qty{\displaystyle\bigcup_{N=1}^\infty \displaystyle\bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu \qty{\displaystyle\bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu(E_M^c) \\ &\coloneqq\sum_{N=1}^\infty \mu(X\setminus E_N) \\ &\overset{N\to\infty}\to 0 .\end{align*}

\todo[inline]{Last step seems wrong!}

Part a:

\(\implies\):

• Suppose \(\chi_{E_n}\to 1\) uniformly, we want to produce an \(N\) such that \(n\geq N \implies x\in E_n\) for all \(x\in X\).
• Take \({\varepsilon}\coloneqq 1/2\). By uniform convergence, for \(N\) large enough, \begin{align*} & \forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} < 1/2 && \forall x\in X\\ &\iff \forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\iff \forall n\geq N \quad x\in E_n && \forall x\in X &\iff \forall n\geq N \quad E_n = X ,\end{align*} where we’ve used that \(E_n \subseteq X\) by definition and this shows \(X \subseteq E_n\). So this \(N\) suffices.

\(\impliedby\):

• Let \({\varepsilon}> 0\) be arbitrary.
• Choose \(N\) such that \(n\geq N \implies X = E_n\). Then \begin{align*} &\forall n\geq N \quad x\in E_n && \forall x\in X \\ &\forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} = 0 < {\varepsilon}&& \forall x\in X ,\end{align*} so \(\chi_{E_n} \to 1\) uniformly.

Part b:

• Define \begin{align*} S &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\chi_{E_k}(x) \to 1}\right\}\\ &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\forall {\varepsilon},\, \exists N\, \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon},\forall k\geq N}\right\}\\ L &\coloneqq\displaystyle\bigcap_{n\geq 0} \displaystyle\bigcup_{k\geq n} \qty{X\setminus E_k} ,\end{align*} so \(S\) is the set where \(f_n\to f\) and \(X\setminus S\) is the exceptional set where \(f_n\not\to f\) doesn’t converge pointwise.

• Claim: \(L = X\setminus S\), so if \(x\in S \iff x\in X\setminus L\).

• Proof of claim: Suppose there exists an \(N\) such that the first line below is true. Then for a fixed \(x\), there are equivalent statements: \begin{align*} &\qquad x \in S \\ &\iff \exists N \text{ s.t. } \forall {\varepsilon}>0,\quad {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon}&& \forall k\geq N \\ &\iff \exists N \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} = 0 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } \chi_{E_k}(x) = 1 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\in E_k && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in X\setminus E_k &&\forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in \displaystyle\bigcup_{k\geq N} X\setminus E_k \\ &{\color{blue} \iff} x\not\in \displaystyle\bigcap_{n\geq 0}\displaystyle\bigcup_{k\geq n} X\setminus E_k \\ &\iff x\not\in L \\ &\iff x\in X\setminus L .\end{align*}

• Proving the iff: \(f_n\to f\) almost everywhere \(\iff \mu(X\setminus S) = 0 \iff \mu(L) = 0\).

• Write \(X\coloneqq[a, b]\),
• Suppose it is not the case that \(f=g\) almost everywhere; then letting \(A\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}f(x) \neq g(x)}\right\}\), we have \(m(A) > 0\).
• Write \begin{align*} A = A_1 {\textstyle\coprod}A_2 \coloneqq\left\{{f > g}\right\} {\textstyle\coprod}\left\{{f < g}\right\} .\end{align*}
• Both \(A_i\) are measurable:
  • Since \(f,g\) are measurable functions, so is \(h\coloneqq f-g\).
  • We can write \begin{align*} A_1 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h > 0 }\right\} = h^{-1}((0, \infty)) \\ A_2 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h < 0 }\right\} = h^{-1}((-\infty, 0)) ,\end{align*} and pullbacks of Borel sets by measurable functions are measurable.
• Then on \(E\), we have \(f(x)>g(x)\) pointwise. This is preserved by monotonicity of the integral, thus \begin{align*} f(x) > g(x) \text{ on } E \implies \int_{E} f(x)\,dx > \int_{E} g(x)\, dx .\end{align*}