\(L^p\) Completeness & Convergence

Proof:

Modifying each \(g_n\) on a null set does not change any integral, so we may assume the monotonicity \(0 \leq g_1 \leq g_2 \leq \cdots\) holds everywhere. Then \(g(x) = \lim_n g_n(x) = \sup_n g_n(x)\) is measurable as the pointwise supremum of measurable functions.

Easy direction (\(\leq\)): Since \(g_n \leq g\) pointwise, every simple function \(q \in S(g_n)\) also satisfies \(q \leq g\), i.e., \(S(g_n) \subseteq S(g)\). Hence \[ \int g_n \, d\mu \;=\; \sup_{q \in S(g_n)} \int q \, d\mu \;\leq\; \sup_{q \in S(g)} \int q \, d\mu \;=\; \int g \, d\mu \] for every \(n\). The sequence \(\bigl(\int g_n \, d\mu\bigr)\) is non-decreasing (same argument applied to \(g_n \leq g_{n+1}\)), so its limit exists in \([0, \infty]\) and \[ \lim_{n \to \infty} \int g_n \, d\mu \;\leq\; \int g \, d\mu. \]

Hard direction (\(\geq\)): We show \(\int q \, d\mu \leq \lim_n \int g_n \, d\mu\) for every simple \(q \in S(g)\); taking the supremum over such \(q\) then yields the desired inequality, by definition of \(\int g \, d\mu\).

Fix \(q = \sum_{i=1}^k a_i \, \chi_{A_i} \in S(g)\), with \(a_i \geq 0\) and \(\{A_i\}\) disjoint measurable. Fix \(\alpha \in (0, 1)\) and define \[ E_n \;=\; \{x \in \Omega : g_n(x) \geq \alpha \, q(x)\}. \] Each \(E_n\) is measurable: on \(\bigcup_i A_i\), \(q\) takes the constant value \(a_i\) on each \(A_i\), so \(E_n \cap A_i = \{g_n \geq \alpha a_i\} \cap A_i\); on the complement \(\Omega \setminus \bigcup_i A_i\), \(q\) vanishes and \(E_n\) contains this complement entirely (since \(g_n \geq 0\)). Hence \[ E_n \;=\; \Bigl(\bigcup_{i=1}^k \bigl(\{g_n \geq \alpha a_i\} \cap A_i\bigr)\Bigr) \cup \Bigl(\Omega \setminus \bigcup_{i=1}^k A_i\Bigr), \] a finite combination of measurable sets, hence measurable. The sequence is increasing because \(g_n\) is. We claim \(\bigcup_n E_n = \Omega\). Indeed, fix \(x \in \Omega\):

• If \(q(x) = 0\), then \(g_n(x) \geq 0 = \alpha q(x)\) for every \(n\), so \(x \in E_1 \subseteq \bigcup E_n\).
• If \(q(x) > 0\), then \(g(x) \geq q(x) > \alpha q(x)\) (strict, since \(\alpha < 1\)), and \(g_n(x) \uparrow g(x)\), so eventually \(g_n(x) \geq \alpha q(x)\), placing \(x\) in some \(E_n\).

On \(E_n\) we have \(g_n \geq \alpha q\), so \[ \int g_n \, d\mu \;\geq\; \int_{E_n} g_n \, d\mu \;\geq\; \alpha \int_{E_n} q \, d\mu \;=\; \alpha \sum_{i=1}^k a_i \, \mu(A_i \cap E_n). \] For each \(i\), the sets \(A_i \cap E_n\) increase to \(A_i\) (since \(E_n \uparrow \Omega\)), so by continuity from below, \(\mu(A_i \cap E_n) \uparrow \mu(A_i)\). The sum has finitely many terms, so we may pass to the limit term-by-term: \[ \lim_{n \to \infty} \int g_n \, d\mu \;\geq\; \alpha \sum_{i=1}^k a_i \, \mu(A_i) \;=\; \alpha \int q \, d\mu. \] This holds for every \(\alpha \in (0, 1)\); letting \(\alpha \to 1^-\) gives \(\lim_n \int g_n \, d\mu \geq \int q \, d\mu\). Taking the supremum over \(q \in S(g)\) yields \(\lim_n \int g_n \, d\mu \geq \int g \, d\mu\), completing the proof.

Proof:

As in MCT, modifying each \(g_n\) on the null set \(N_n = \{g_n < 0\}\) does not affect any integral; on the complement of \(N = \bigcup_n N_n\) (itself null), all \(g_n\) are non-negative. Working modulo \(N\), we may assume \(g_n \geq 0\) everywhere. For each \(n\), set \(h_n(x) = \inf_{k \geq n} g_k(x)\). Then \(0 \leq h_1 \leq h_2 \leq \cdots\) (the infimum over a smaller index set is larger), \(h_n\) is measurable as the pointwise infimum of countably many measurable functions, and by definition \[ \lim_{n \to \infty} h_n(x) \;=\; \sup_n \inf_{k \geq n} g_k(x) \;=\; \liminf_{n \to \infty} g_n(x). \] Applying MCT to \((h_n)\): \[ \int \liminf_{n \to \infty} g_n \, d\mu \;=\; \lim_{n \to \infty} \int h_n \, d\mu. \] Since \(h_n \leq g_n\) pointwise (the infimum is at most each member), monotonicity of the integral gives \(\int h_n \, d\mu \leq \int g_n \, d\mu\) for every \(n\). Taking \(\liminf\) on both sides and using \(\lim \int h_n \, d\mu = \liminf \int h_n \, d\mu\) (the limit exists), \[ \int \liminf_{n \to \infty} g_n \, d\mu \;=\; \liminf_{n \to \infty} \int h_n \, d\mu \;\leq\; \liminf_{n \to \infty} \int g_n \, d\mu. \]

Proof:

Modifying on a null set, assume the convergence \(f_n \to f\) and the bound \(|f_n| \leq h\) hold everywhere. Then \(f\) is measurable as the pointwise limit of measurable functions, and \(|f| \leq h\) everywhere gives \(\int |f| \, d\mu \leq \int h \, d\mu < \infty\), hence \(f \in L^1\). For brevity we treat the real-valued case; the complex case follows by splitting \(f_n = \mathrm{Re}(f_n) + i\,\mathrm{Im}(f_n)\) and applying the real result to each part, noting that \(|\mathrm{Re}(f_n)|, |\mathrm{Im}(f_n)| \leq |f_n| \leq h\) and both parts converge pointwise.

The sequences \(h + f_n\) and \(h - f_n\) are non-negative (since \(|f_n| \leq h\)) and converge pointwise to \(h + f\) and \(h - f\) respectively. Note that \(|f_n| \leq h \in L^1\) gives \(\bigl|\int f_n \, d\mu\bigr| \leq \int h \, d\mu < \infty\), so the real sequence \(\bigl(\int f_n \, d\mu\bigr)\) is bounded, and both \(\liminf\) and \(\limsup\) are finite real numbers — in particular, the identity \(\liminf(-a_n) = -\limsup(a_n)\) applies without \(\infty - \infty\) ambiguity. Apply Fatou's lemma to each: \[ \int (h + f) \, d\mu \;=\; \int \liminf_{n} (h + f_n) \, d\mu \;\leq\; \liminf_{n} \int (h + f_n) \, d\mu \;=\; \int h \, d\mu + \liminf_{n} \int f_n \, d\mu, \] \[ \int (h - f) \, d\mu \;=\; \int \liminf_{n} (h - f_n) \, d\mu \;\leq\; \liminf_{n} \int (h - f_n) \, d\mu \;=\; \int h \, d\mu - \limsup_{n} \int f_n \, d\mu, \] where the last equality uses \(\liminf(-a_n) = -\limsup(a_n)\). On each left-hand side, expand \(\int (h \pm f) \, d\mu = \int h \, d\mu \pm \int f \, d\mu\) (integral linearity, valid since \(h, f \in L^1\)) and subtract \(\int h \, d\mu\) (finite, hence cancellable): \[ \int f \, d\mu \;\leq\; \liminf_{n} \int f_n \, d\mu, \qquad -\int f \, d\mu \;\leq\; -\limsup_{n} \int f_n \, d\mu. \] The second rearranges to \(\limsup_n \int f_n \, d\mu \leq \int f \, d\mu\). Combining with the first, \[ \limsup_{n} \int f_n \, d\mu \;\leq\; \int f \, d\mu \;\leq\; \liminf_{n} \int f_n \, d\mu. \] Since \(\liminf \leq \limsup\) always, all three quantities coincide; the common value is \(\lim_n \int f_n \, d\mu\), which equals \(\int f \, d\mu\).

Proof:

(\(\Rightarrow\)) Assume \(\mathcal{X}\) is complete and \(\sum \|h_k\| < \infty\). For \(M > N\), the partial sums satisfy \(\bigl\|\sum_{k=1}^{M} h_k - \sum_{k=1}^{N} h_k\bigr\| = \bigl\|\sum_{k=N+1}^{M} h_k\bigr\| \leq \sum_{k=N+1}^{M} \|h_k\| \to 0\) as \(N, M \to \infty\), since the tail of a convergent series tends to zero. Hence the partial sums form a Cauchy sequence and converge by completeness.

(\(\Leftarrow\)) Assume every absolutely summable series converges. Let \((x_n)\) be a Cauchy sequence in \(\mathcal{X}\). Extract a fast subsequence \(x_{n_1}, x_{n_2}, \ldots\) with \(\|x_{n_{k+1}} - x_{n_k}\| < 2^{-k}\). Setting \(h_k = x_{n_{k+1}} - x_{n_k}\), we have \(\sum \|h_k\| < \sum 2^{-k} = 1 < \infty\), so by hypothesis the series \(\sum h_k\) converges. Its partial sums telescope to \(x_{n_{K+1}} - x_{n_1}\), so the subsequence \((x_{n_k})\) converges to some \(x\). To upgrade convergence of the subsequence to convergence of the full sequence, we use the standard \(\epsilon/2\) argument: given \(\epsilon > 0\), choose \(N_0\) so that \(\|x_m - x_n\| < \epsilon/2\) for \(m, n \geq N_0\), and \(K\) so that \(n_K \geq N_0\) and \(\|x_{n_K} - x\| < \epsilon/2\); then for all \(n \geq N_0\), \(\|x_n - x\| \leq \|x_n - x_{n_K}\| + \|x_{n_K} - x\| < \epsilon\). Hence \(x_n \to x\). (This same lifting argument reappears as Step 5 of the Riesz-Fischer proof below.)

Proof:

Step 1 — Extract a fast subsequence.
Since \((f_n)\) is Cauchy, for each \(k \geq 1\) there exists a threshold index \(N_k\) such that \(\|f_m - f_n\|_p < 2^{-k}\) for all \(m, n \geq N_k\); replacing \(N_k\) by \(\max(N_1, \dots, N_k)\) if necessary, we may assume \(N_1 \leq N_2 \leq \cdots\). Construct \((n_k)\) inductively: pick \(n_1 \geq N_1\), and given \(n_k\), pick \(n_{k+1} > n_k\) with \(n_{k+1} \geq N_{k+1}\) (always possible since the Cauchy thresholds impose only a lower bound on indices). Since \(n_k, n_{k+1} \geq N_k\) (using the monotonicity of \((N_k)\)), the resulting subsequence \((f_{n_k})\) is strictly index-increasing and satisfies \[ \|f_{n_{k+1}} - f_{n_k}\|_p \;<\; 2^{-k} \quad \text{for all } k \geq 1. \] In particular, the "differences" \(h_k = f_{n_{k+1}} - f_{n_k}\) satisfy \(\sum_{k=1}^{\infty} \|h_k\|_p < \sum_{k=1}^{\infty} 2^{-k} = 1 < \infty\).

Step 2 — Construct a dominating function via MCT.
Define the partial sums of absolute values: \[ G_N(x) \;=\; |f_{n_1}(x)| + \sum_{k=1}^{N} |f_{n_{k+1}}(x) - f_{n_k}(x)|. \] The sequence \((G_N)\) is nonnegative, pointwise increasing, and measurable (as finite sums of measurable functions). Since \(t \mapsto t^p\) is continuous and nondecreasing on \([0, \infty)\) for any \(p > 0\) (a fortiori for \(p \geq 1\)), the sequence \((G_N^p)\) is also nonnegative, pointwise increasing, and measurable. Applying the Monotone Convergence Theorem to \((G_N^p)\): \[ \int_\Omega G^p \, d\mu \;=\; \lim_{N \to \infty} \int_\Omega G_N^p \, d\mu, \] where \(G(x) = \lim_{N \to \infty} G_N(x)\). Taking \(p\)-th roots (the continuous map \(t \mapsto t^{1/p}\) on \([0, \infty]\) preserves limits), \(\|G\|_p = \lim_{N \to \infty} \|G_N\|_p\). Each \(|h_k|\) lies in \(L^p\) (since \(h_k \in L^p\) and \(\||h_k|\|_p = \|h_k\|_p\)), so applying Minkowski's inequality finitely many times gives \[ \|G_N\|_p \;\leq\; \|f_{n_1}\|_p + \sum_{k=1}^{N} \|h_k\|_p \;\leq\; \|f_{n_1}\|_p + 1. \] Therefore \(\|G\|_p \leq \|f_{n_1}\|_p + 1 < \infty\), which means \(G \in L^p\). In particular, since \(\int G^p \, d\mu < \infty\) and \(G \geq 0\), we have \(G(x) < \infty\) for a.e. \(x\).

Step 3 — Obtain a pointwise limit.
Fix any \(x\) with \(G(x) < \infty\) (which holds for a.e. \(x\), as established in Step 2). From the definition of \(G\) and \(G_N\), \[ \sum_{k=1}^{\infty} |f_{n_{k+1}}(x) - f_{n_k}(x)| \;=\; G(x) - |f_{n_1}(x)| \;<\; \infty, \] so the series \(\sum_{k=1}^{\infty} (f_{n_{k+1}}(x) - f_{n_k}(x))\) is absolutely convergent in \(\mathbb{F}\), hence convergent (since \(\mathbb{F}\) is complete). By telescoping, \(\sum_{k=1}^{K} h_k(x) = f_{n_{K+1}}(x) - f_{n_1}(x)\), so \(f_{n_{K+1}}(x) = f_{n_1}(x) + \sum_{k=1}^{K} h_k(x)\) converges as \(K \to \infty\); reindexing, the subsequence \((f_{n_K}(x))\) itself converges. Define \[ f(x) \;=\; \lim_{K \to \infty} f_{n_K}(x). \] (On the measure-zero set where \(G(x) = \infty\), we set \(f(x) = 0\), say; since this set is null, the resulting a.e.-equivalence class is independent of the choice.)

We show \(|f_{n_K}(x)| \leq G(x)\) for every \(K\) and every \(x\) with \(G(x) < \infty\). Indeed, by the same telescoping plus the triangle inequality, \[ |f_{n_K}(x)| \;=\; \biggl| f_{n_1}(x) + \sum_{k=1}^{K-1} h_k(x) \biggr| \;\leq\; |f_{n_1}(x)| + \sum_{k=1}^{K-1} |h_k(x)| \;\leq\; |f_{n_1}(x)| + \sum_{k=1}^{\infty} |h_k(x)| \;=\; G(x). \] Passing to the limit \(K \to \infty\) — using \(f_{n_K}(x) \to f(x)\) and continuity of \(|\cdot|\) — gives \(|f(x)| \leq G(x)\) for a.e. \(x\) (namely, on \(\{G < \infty\}\)). The function \(f\) is measurable as the a.e.-pointwise limit of measurable functions (extended by \(0\) on the null set where \(G = \infty\)), and \(G \in L^p\) together with \(|f|^p \leq G^p\) a.e. yields, by monotonicity of the integral, \(\int |f|^p \, d\mu \leq \int G^p \, d\mu < \infty\), so \(f \in L^p\).

Step 4 — Prove \(L^p\) convergence of the subsequence.
From Step 3, \(f_{n_K}(x) \to f(x)\) for a.e. \(x\); since \(t \mapsto |t|^p\) is continuous on \(\mathbb{F}\), this gives \(|f_{n_K} - f|^p \to 0\) a.e. Furthermore, using \(|f_{n_K}|, |f| \leq G\) a.e. (Step 3), \[ |f_{n_K}(x) - f(x)|^p \;\leq\; \bigl(|f_{n_K}(x)| + |f(x)|\bigr)^p \;\leq\; (G(x) + G(x))^p \;=\; (2G(x))^p, \] and \(\int (2G)^p \, d\mu = 2^p \int G^p \, d\mu < \infty\), so \((2G)^p \in L^1\) is an admissible dominator. By the Dominated Convergence Theorem: \[ \|f_{n_K} - f\|_p^p \;=\; \int |f_{n_K} - f|^p \, d\mu \;\to\; 0 \quad \text{as } K \to \infty. \] Since \(t \mapsto t^{1/p}\) is continuous on \([0, \infty)\) with \(t^{1/p} = 0 \iff t = 0\), this gives \(\|f_{n_K} - f\|_p \to 0\).

Step 5 — Lift from the subsequence to the full sequence.
(This is the same lifting argument as in Lemma: Absolute Summability Criterion (\(\Leftarrow\)), applied concretely to \((f_n)\).) We now know \(f_{n_K} \to f\) in \(L^p\). To show \(f_n \to f\) in \(L^p\), we use the fact that \((f_n)\) is Cauchy. Fix \(\epsilon > 0\) and choose \(N_0\) such that \(\|f_m - f_n\|_p < \epsilon/2\) for all \(m, n \geq N_0\). Since \(n_K \to \infty\) and \(\|f_{n_K} - f\|_p \to 0\), we can pick a single \(K\) satisfying both \(n_K \geq N_0\) and \(\|f_{n_K} - f\|_p < \epsilon/2\). Then for all \(n \geq N_0\), the Cauchy condition applies to the pair \((n, n_K)\), giving \[ \|f_n - f\|_p \;\leq\; \|f_n - f_{n_K}\|_p + \|f_{n_K} - f\|_p \;<\; \frac{\epsilon}{2} + \frac{\epsilon}{2} \;=\; \epsilon. \] Hence \(f_n \to f\) in \(L^p\), completing the proof for \(1 \leq p < \infty\).

Proof:

Case \(1 \leq p < \infty\): Since \((f_n)\) converges in \(L^p\), it is Cauchy. Apply Steps 1-3 of the Riesz-Fischer proof to extract a subsequence \((f_{n_k})\) and produce an a.e. pointwise limit \(\tilde f \in L^p\) with \(f_{n_k}(x) \to \tilde f(x)\) for a.e. \(x\); Step 4 further gives \(f_{n_k} \to \tilde f\) in \(L^p\). On the other hand, by hypothesis \(f_n \to f\) in \(L^p\), so the subsequence also converges in \(L^p\) to \(f\). The \(L^p\) limit is unique (if \(g_n \to g\) and \(g_n \to g'\) in \(L^p\), then \(\|g - g'\|_p \leq \|g - g_n\|_p + \|g_n - g'\|_p \to 0\), so \(\|g - g'\|_p = 0\), i.e., \(g = g'\) a.e.), so \(\tilde f = f\) a.e. Therefore \(f_{n_k}(x) \to f(x)\) for a.e. \(x\), as claimed.

Case \(p = \infty\): The argument in the Case \(p = \infty\) section above (applied to \((f_n)\), which is Cauchy because it converges) directly produces a function \(\tilde f\) with \(f_n(x) \to \tilde f(x)\) for every \(x \notin E\), where \(E\) is a null set. The same uniqueness argument as above identifies \(\tilde f = f\) a.e. Hence the full sequence — not merely a subsequence — converges a.e. to \(f\), and the corollary holds trivially.

Proof of equivalence:

The forward direction is immediate, since \(\sup_{x \notin E}|f_n - f|\) is an admissible essential bound for \(|f_n - f|\), so \(\|f_n - f\|_\infty \leq \sup_{x \notin E}|f_n - f| \to 0\). The reverse uses Lemma: Essential Supremum Is Attained: for each \(n\), \(|f_n - f| \leq \|f_n - f\|_\infty\) outside a null set \(E_n\); setting \(E = \bigcup_n E_n\) (still null as a countable union), the inequality \(|f_n(x) - f(x)| \leq \|f_n - f\|_\infty\) holds for every \(x \notin E\) and every \(n\), giving \(\sup_{x \notin E}|f_n - f| \leq \|f_n - f\|_\infty \to 0\).

Proof:

Let \(N\) be the null set outside which \(f_n(x) \to f(x)\). Fix \(\epsilon > 0\) and define \[ B_n \;=\; \bigcup_{k \geq n} \{|f_k - f| > \epsilon\}. \] The sequence \((B_n)\) is decreasing (\(B_{n+1} \subseteq B_n\)). We claim \(\bigcap_n B_n \subseteq N\): for \(x \notin N\), there exists \(n_0 = n_0(x, \epsilon)\) such that \(|f_k(x) - f(x)| \leq \epsilon\) for all \(k \geq n_0\), so \(x \notin B_{n_0}\) and hence \(x \notin \bigcap_n B_n\). Therefore \(\mu\bigl(\bigcap_n B_n\bigr) \leq \mu(N) = 0\). Since \(\mu(B_1) \leq \mu(\Omega) < \infty\), continuity of measure from above applies, giving \(\mu(B_n) \to 0\). Since \(\{|f_n - f| > \epsilon\} \subseteq B_n\), we conclude \(\mu(\{|f_n - f| > \epsilon\}) \to 0\), i.e., \(f_n \to f\) in measure.

Proof of (1) — \(1 \leq p < \infty\) case:

This follows from the Chebyshev-Markov inequality, which we briefly justify here for completeness: for any non-negative measurable \(g\) and any \(t > 0\), \[ \int_\Omega g \, d\mu \;\geq\; \int_{\{g > t\}} g \, d\mu \;\geq\; t \cdot \mu(\{g > t\}), \] so \(\mu(\{g > t\}) \leq t^{-1} \int g \, d\mu\). Applying this with \(g = |f_n - f|^p\) and \(t = \epsilon^p\), \[ \mu\bigl(\{|f_n - f| > \epsilon\}\bigr) \;=\; \mu\bigl(\{|f_n - f|^p > \epsilon^p\}\bigr) \;\leq\; \frac{1}{\epsilon^p} \int_\Omega |f_n - f|^p \, d\mu \;=\; \frac{\|f_n - f\|_p^p}{\epsilon^p}. \] If \(\|f_n - f\|_p \to 0\), the right side tends to zero, so \(f_n \to f\) in measure.

The case \(p = \infty\): If \(\|f_n - f\|_\infty \to 0\), then by Lemma: Essential Supremum Is Attained, for any \(\epsilon > 0\), once \(n\) is large enough that \(\|f_n - f\|_\infty < \epsilon\), the set \(\{|f_n - f| > \epsilon\}\) is contained in the null exceptional set of the lemma. Hence \(\mu(\{|f_n - f| > \epsilon\}) = 0\) for all sufficiently large \(n\), which is even stronger than required.

Proof of (4) — convergence in measure \(\Rightarrow\) subsequence converges a.e.:

Suppose \(f_n \to f\) in measure. For each \(k \geq 1\), \(\mu(\{|f_n - f| > 2^{-k}\}) \to 0\) as \(n \to \infty\), so there exists \(N_k\) with \(\mu(\{|f_n - f| > 2^{-k}\}) < 2^{-k}\) for all \(n \geq N_k\); replacing \(N_k\) by \(\max(N_1, \dots, N_k)\) if necessary, we may assume \(N_1 \leq N_2 \leq \cdots\). Pick \(n_1 \geq N_1\), and given \(n_k\), pick \(n_{k+1} > n_k\) with \(n_{k+1} \geq N_{k+1}\). Then \(n_k \geq N_k\) for every \(k\), so \[ \mu(A_k) \;<\; 2^{-k}, \qquad \text{where } A_k = \{|f_{n_k} - f| > 2^{-k}\}. \] Define the "tail" sets \(B_m = \bigcup_{k \geq m} A_k\). Writing \(B_m\) as a disjoint union \(B_m = \bigcup_{k \geq m} \bigl(A_k \setminus \bigcup_{j < k} A_j\bigr)\) (each piece a subset of \(A_k\)) and applying countable additivity with monotonicity yields the \(\sigma\)-subadditivity bound \(\mu(B_m) \leq \sum_{k \geq m} \mu(A_k) < \sum_{k \geq m} 2^{-k} = 2^{-m+1}\), which tends to \(0\) as \(m \to \infty\). Set \(B = \bigcap_{m=1}^\infty B_m\); since \(\mu(B) \leq \mu(B_m)\) for every \(m\), we have \(\mu(B) = 0\), so \(B\) is null.

For \(x \notin B\), there exists \(m_0\) (depending on \(x\)) with \(x \notin B_{m_0}\), i.e., \(x \notin A_k\) for every \(k \geq m_0\); equivalently, \(|f_{n_k}(x) - f(x)| \leq 2^{-k}\) for every \(k \geq m_0\), so \(f_{n_k}(x) \to f(x)\). This holds for every \(x \notin B\), giving \(f_{n_k} \to f\) a.e.

Counterexample (\(1 \leq p < \infty\)):

Consider the interval \([0, 1]\) with Lebesgue measure. We construct a sequence of indicator functions that converges to zero in \(L^p\) but does not converge at any point.

Enumerate the dyadic-style intervals in blocks: the \(j\)-th block consists of the \(j\) intervals \([0, 1/j], [1/j, 2/j], \ldots, [(j-1)/j, 1]\), each of width \(1/j\). (The intervals within a block share endpoints at the points \(k/j\), \(k = 1, \ldots, j-1\); this overlap occurs on a set of measure zero and does not affect any of the \(L^p\) computations below.) Listing the blocks consecutively for \(j = 1, 2, 3, \ldots\) gives the sequence \[ \underbrace{[0,1]}_{j=1},\;\; \underbrace{[0, \tfrac{1}{2}],\, [\tfrac{1}{2}, 1]}_{j=2},\;\; \underbrace{[0, \tfrac{1}{3}],\, [\tfrac{1}{3}, \tfrac{2}{3}],\, [\tfrac{2}{3}, 1]}_{j=3},\;\; \underbrace{[0, \tfrac{1}{4}], \ldots}_{j=4},\;\; \ldots \] Let \(f_n = \chi_{I_n}\) where \(I_n\) is the \(n\)-th interval. The \(j\)-th block ends at index \(j(j+1)/2\), so an index \(n\) belongs to block \(j\) precisely when \(j(j-1)/2 < n \leq j(j+1)/2\); in particular, \(j \to \infty\) as \(n \to \infty\), and the corresponding interval has width \(|I_n| = 1/j\). Therefore \(\|f_n\|_p = |I_n|^{1/p} = j^{-1/p} \to 0\), so \(f_n \to 0\) in \(L^p\).

However, fix any \(x \in [0, 1]\). Within the \(j\)-th block, \(x\) lies in at least one of the \(j\) intervals (the block tiles \([0,1]\), with at most two intervals overlapping at the rational endpoints \(k/j\)), so \(f_n(x) = 1\) for at least one \(n\) in the block. For \(j \geq 2\), \(x\) lies in at most two intervals of the block, so \(f_n(x) = 0\) for at least \(j - 2 \geq 0\) values of \(n\) in the block — and \(j - 2 \geq 1\) for \(j \geq 3\). Letting \(j \to \infty\), the value \(1\) is attained infinitely often and the value \(0\) is attained infinitely often. Therefore \(f_n(x)\) fails to converge at every \(x \in [0, 1]\); the sequence does not converge a.e. — in fact, it does not converge at any point at all.

Proof:

Modifying on a null set, both \(|f_n| \leq h\) and \(f_n \to f\) hold outside a single null set (the union of the individual null sets is null as a countable union). Define \(f\) to be the a.e. pointwise limit on the good set, extended by zero on the exceptional null set; then \(f\) is measurable as the pointwise limit of measurable functions on the good set. On the good set, \(|f(x)| = \lim_n |f_n(x)| \leq h(x)\), so \(|f| \leq h\) a.e. Since \(p \geq 1\) and \(t \mapsto t^p\) is non-decreasing on \([0, \infty)\), this gives \(|f|^p \leq h^p\) a.e. By monotonicity of the integral, \(\int |f|^p \, d\mu \leq \int h^p \, d\mu < \infty\), so \(f \in L^p\).

Now consider \(|f_n - f|^p \leq (|f_n| + |f|)^p \leq (h + h)^p = (2h)^p\) a.e., and \(\int (2h)^p \, d\mu = 2^p \int h^p \, d\mu < \infty\), so \((2h)^p \in L^1\) is an admissible dominator. Since \(t \mapsto |t|^p\) is continuous on \(\mathbb{F}\) and \(f_n \to f\) a.e., we have \(|f_n - f|^p \to 0\) a.e. Applying the standard Dominated Convergence Theorem to the sequence \(|f_n - f|^p\) with dominator \((2h)^p\) yields \(\int |f_n - f|^p \, d\mu \to 0\), and since \(t \mapsto t^{1/p}\) is continuous on \([0, \infty)\) with \(t^{1/p} = 0 \iff t = 0\), this gives \(\|f_n - f\|_p \to 0\).