Function Sequences #

The ring of finite measure #

Suppose $(\Omega, \A, \mu)$ is a measure space.

• Define $\A^<: =\{A \in \A: \mu(A) < \infty\}$. Clearly $\A^<$ is a ring on $\Omega$.

Markov inequality #

Suppose $(\Omega, \A, \mu)$ is a measure space. Suppose $f \in \L(\Omega \to [0, \infty))$.

Now we warp the values of $f$ by increasing $\theta: [0, \infty) \to [0, \infty)$. Suppose $\mu[\theta(f)]$ exists.

• Then for any $\epsilon > 0$: $$ \mu[\theta(f)] \ge \mu[\theta(f)1 _ {\theta(f) \ge \theta(\epsilon)}] \ge \mu[\theta(\epsilon)1 _ {\theta(f) \ge \theta(\epsilon)}] = \theta(\epsilon)\mu\{f \ge \epsilon\} $$
• Let $\theta(x) = x^p$ for $p \in (0, \infty)$. And $f \in \L^p(\Omega \to [0, \infty])$. Then $\epsilon^p \mu\{f \ge \epsilon\} \le \mu f^p$.
  • Apply to $\L^p$ space, this gives $\mu\{|f| \ge \epsilon\} \le \epsilon^{-p} \|f\| _ p^p$.

Continuity of metric #

Suppose $(E, d)$ is a (extended) metric space. $d(x, y): E^2 \to [0, \infty]$ is a continuous function.

• $E^2$ is equipped with the product metric.
• $d$ is continuous iff for all sequence $\p{x _ n, y _ n} \to (x, y)$ in $E^2$, $d(x _ n, y _ n) \to d(x, y)$.
• When $d(x, y) < \infty$.
  • $d(x _ n, y _ n) \le d(x _ n, x) + d(x, y) + d(y, y _ n)$.
  • So $|d(x _ n, y _ n) - d(x, y)| \le d(x _ n, x) + d(y _ n, y) \le d((x _ n, y _ n), (x, y))$.
  • So $d(x _ n, y _ n) \to d(x, y)$.
• When $d(x, y) = \infty$. $d(x _ n, y _ n) \to \infty$.

Suppose $f, g \in \L(\Omega \to E, \A)$.

• Then $\omega \mapsto (f(\omega), g(\omega)) \in \L(\Omega \to E^2, \A/\B(E^2))$.
• Since $d: \L(E^2 \to [0, \infty], \B(E^2))$
• So $H(\omega) = d(f(\omega), g(\omega)) \in \L(\Omega \to [0, \infty], \A)$.

Uniformly bounded #

Suppose $\p{f _ n} _ {n=1}^\infty: \Omega \to E$ where $E$ is a metric space.

$\p{f _ n}$ is uniformly bounded if $\diam \cup _ n f _ n[\Omega] < \infty$.

Basic modes of convergence #

Suppose $f, \p{f _ n} _ {n=1}^\infty: \Omega \to E$ where $E$ is a metric space.

1. (Pointwise) $f _ n$ converge to $f$ pointwise on $X \subseteq \Omega$, denoted as $f _ n \to _ X f$.
2. (Uniform) $\forall \epsilon > 0, \exists N \in \Z, \forall n \ge N, \forall x \in X: d(f _ n(x), f(x)) < \epsilon$.
   • Denoted by $f _ n \rightrightarrows _ X f$, $f _ n$ converge to $f$ uniformly on $X$.
   • $f _ n \rightrightarrows _ X f$ then $f _ n \to _ X f$.

Let $X$ be a topological space. Let $M$ be a metric space. Let $f _ n: X \to M$ and $f: X \to M$.

Suppose $\Omega$ is a topological space. And $X \subseteq \Omega$.

3. (Locally uniform) $f _ n$ converge to $f$ locally uniformly if $\forall x \in X, \exists U \in N(x): f _ n \rightrightarrows _ U f$.
4. (Bounded uniform) $f _ n$ converges bounded uniformly to $f$ on $X$ if $\forall E \in \P _ {\operatorname{bounded}}(X): f _ n \rightrightarrows _ E f$.
   • Bounded uniformly convergence implies locally uniformly convergence.

Limit of uniformly continuous functions #

Suppose $S$ is a topological space. And $f, \p{f _ n} _ {n=1}^\infty: S \to E$ where $E$ is a metric space.

Suppose $f _ n \rightrightarrows _ S f$. And $f _ n$ are continuous at $c \in S$, then $f$ is continuous at $c \in S$.

• Consider any $\epsilon > 0$.
• There exists $N$ where for all $n \ge N$ we have $d(f _ n(x), f(x)) < \epsilon / 3$.
• Now pick any $n \ge N$. There exists open neighborhood $O$ of $c$ where $f _ n(O) \subseteq B(f _ n(c), \epsilon / 3)$.
• For any $x \in O$, consider chain $f(x) \to f _ n(x) \to f _ n(c) \to f(c)$, each link has error at most $\epsilon / 3$.

$$ \lim _ {x \to c} \lim _ {n \to \infty} f _ {n}(x)=\lim _ {n \to \infty} \lim _ {x \to c} f _ {n}(x) $$

Cauchy condition for uniform convergence #

Suppose $\p{f _ n} _ {n=1}^\infty: S \to E$ where $E$ is a complete metric space.

Then $f _ n \rightrightarrows _ S f$ to some $f: S \to E$ if and only if $$ \forall \epsilon > 0, \exists N \in \Z, \forall x \in S, \forall m, n > N: d(f _ m(x), f _ n(x)) < \epsilon $$

Function Series #

Function series #

Suppose $\p{f _ n} _ {n=1}^\infty: S \to E$ where $E$ is a normed space over $\bF$.

Define the partial sum on $S$ as $s _ n(x) = \sum _ {k=1}^n f _ k(x)$.

• Suppose $s _ n \to _ S f$, series $\sum f _ n$ is convergent on $S$. Denoted as $\sum f _ n \to _ S f$.
• Suppose $s _ n \rightrightarrows _ S f$, series $\sum f _ n$ is uniformly convergent on $S$. Denoted as $\sum f _ n \rightrightarrows _ S f$.

Weierstrass M-test for uniform convergence #

Suppose $\p{f _ n} _ {n=1}^\infty: S \to E$ where $E$ is a normed space over $\bF$.

Then $\sum f _ k$ converges uniformly on $S$. If

• $M _ n > 0$ and $\sum M _ n$ converges.
• And $\forall n \in \N, \forall x\in S: 0 \le \n{f _ n(x)} \le M _ n$.

Dirichlet's Test for Series #

• Suppose $f _ n: S \subseteq \R \to \C$, suppose $F _ n(x) = \sum _ {k=1}^n f _ k(x)$.
• Suppose $F _ n$ is uniformly bounded on $S$.
• Suppose $g _ n : S \to [0, \infty)$. $g _ {n+1}(x) \le g _ n(x)$.
• Suppose $g _ n \rightrightarrows 0$ on $S$.

Then $\sum f _ n(x) g _ n(x)$ converges uniformly on $S$.

Abel's Test for Series #

• Suppose $f _ n: S \subseteq \R \to \C$, suppose $F _ n(x) = \sum _ {k=1}^n f _ k(x)$.
• Suppose $F _ n$ is uniformly convergent on $S$.
• Suppose $g _ n: S \subseteq \R \to \R$, $g _ {n+1}(x) \le g _ n(x)$.
• Suppose $g _ n$ is uniformly bounded on $S$.

Then $\sum f _ n(x) g _ n(x)$ converges uniformly on $S$.

More Modes of Convergence #

Modes of convergence #

Suppose $(\Omega, \A, \mu)$ is a measure space. And $f, (f _ n) _ {n = 1}^\infty \in \L(\Omega \to E)$.

3. (Essentially pointwise) $f _ n \to f$ $\mu$-a.e. if $f _ n \to _ {\Omega \backslash N} f$ where $\mu(N) = 0$.
   • $f _ n \to f$ $\mu$-a.e. then $f _ n \to f$ in $\mu$ locally (see below).
     • Suppose $A \in \A^<$. And $\epsilon > 0$.
     • Define $D _ {n, \epsilon}: \{\omega \in A: \exists m > n: d(f, f _ m) > \epsilon\}$.
     • $\mu _ A\{d(f, f _ n) > \epsilon\} \le \mu _ A(D _ {n, \epsilon}) \to 0$. So $\mu\{d(f, f _ n) > \epsilon\} \to 0$.
4. (Essentially uniform) $f _ n \rightrightarrows f$ $\mu$-a.e. if $f _ n\rightrightarrows _ {\Omega \backslash N} f$ where $\mu (N) = 0$.
5. (Almost uniform) $f _ n \rightrightarrows f$ in $\mu$ if $\forall \epsilon > 0, \exists E _ \epsilon \in \A: \mu(E _ \epsilon) \le \epsilon \land f _ n \rightrightarrows _ {\Omega \backslash E _ \epsilon} f$.
   • $f _ n \rightrightarrows f$ in $\mu$ then $f _ n \to f$ in $\mu$ globally.
     • Suppose $f _ n \to f$ in $\mu$ globally is false. $\exists \epsilon > 0: \mu\{d(f, f _ n) > \epsilon\} \not \to 0$.
     • $\exists \delta > 0, \forall N \in \N, \exists n > N: \mu\{d(f, f _ n) > \epsilon\} > \delta$.
     • By $f _ n \rightrightarrows f$ in $\mu$, $\exists E _ \delta \in \A: \mu(E _ \delta) \le \delta \land f _ n \rightrightarrows _ {\Omega - E _ \delta} f$.
     • So $\exists N _ \delta \in \N, \forall n > N: \mu\{d(f, f _ n) > \epsilon\} \le \delta$. Contradiction!
   • $f _ n \rightrightarrows f$ in $\mu$ then $f _ n \to f$ $\mu$-a.e.
     • $E = \cap _ {n = 1}^\infty E _ {1/n}$. Then $f _ n \to _ {\Omega \backslash E} f$. And $\mu E = 0$.
6. (Locally in measure) $f _ n \to f$ in $\mu$ if $\forall A \in \A^<,\forall \epsilon > 0: \mu _ A\{d(f, f _ n) > \epsilon\} \to 0$.
7. (Globally in measure) $f _ n \to f$ in $\mu$ globally if $\forall \epsilon > 0: \mu\{d(f, f _ n) > \epsilon\} \to 0$.
   • $f _ n \to f$ in $\mu$ globally then $f _ n \to f$ in $\mu$ locally.
   • Suppose $\mu$ is finite, then $f _ n \to f$ in $\mu$ globally iff locally.

Suppose $f, g, (f _ n) _ {n = 1}^\infty \in \L^p(\Omega \to \symbf F, \mu)$ where $p \in (0, \infty]$.

8. (In $\L _ p$ norm) $f _ n \to f$ in $\L _ p$ if $\|f _ n - f\| _ p \to 0$.
   • For $p \ge 1$, $f _ n \to f$ in $\L _ p$ implies $\mu f _ n \to \mu f$.
     • $\n{f _ n - f} _ 1 \le \n{f _ n - f} _ p \n{1} _ {q}$ by Hölder's inequality.
     • Therefore $f _ n \to f$ in $\L _ 1$ norm.
     • Notice that $|\mu f _ n - \mu f| \le \mu |f _ n - f| \to 0$.
   • $f _ n \to f$ in $\L _ \infty$ iff $f _ n \rightrightarrows f$ $\mu$-a.e.
   • $f _ n \to f$ in $\L _ p$ then $f _ n \to f$ in $\mu$ globally.
     • For $p < \infty$, by (Markov) $\mu\{|f _ n - f| \ge \epsilon\} \le \epsilon^{-p}\|f _ n - f\| _ p^p \to 0$.

Counter-examples #

• (Escape to horizontal infinity) $f _ n = 1 _ {[n, n + 1]}$. $f _ n \to 0$ ($\mu$.a.e.) but not in any other modes.
• (Escape to width infinity) $f _ n = 1 _ {[0, n]}/n$. $f _ n \to 0$ in all modes but $\L _ 1$ norm.
• (Escape to vertical infinity) $f _ n = n 1 _ {[1/n, 2/n]}$.
  • $f _ n \to 0$ pointwise, and almost uniformly.
  • Not converge uniformly or in any norm.
• (Typewriter sequence) $f _ n = 1 _ {[(n - 2^k)/2^k, (n - 2^k + 1)/2^k]}$.
  • $f _ n \to 0$ in $\mu$ globally and in finite-norm.
  • Not converge pointwise $\mu$.a.e. or almost uniformly.

Egorov's Theorem #

Suppose $(\Omega, \A, \mu)$ is a finite measure space. Suppose $(E, d)$ is a metric space.

Suppose $(f _ n) _ {n = 1}^\infty \in \L(\Omega \to E)$ and $f _ n \to f$ $\mu$-a.e. then $f _ n \rightrightarrows f$ in $\mu$.

• For $N, m \in \N^+$ define $E _ {N, m} = \{x \in X: \exists n \ge N: f _ n(x) \notin B(f(x), 1/m)\} \in \A$.
• Since $f _ n \to f$ $\mu$-a.e. then $\mu (\cap _ {N=1}^\infty E _ {N, m}) = 0$. So $\lim _ {N \to \infty} \mu(E _ {N, m}) = 0$.
• For any $m$ there is a $N _ m$ where $\mu(E _ {N _ m, m}) < \epsilon/2^m$.
• $\forall m \in \N^+, \exists N _ m \in \N^+: \mu(E _ {N _ m, m}) < \epsilon / 2^m$. Let $E _ \epsilon = \cup _ {m=1}^\infty E _ {N _ m, m}$. $\mu E _ \epsilon < \epsilon$.
• Clearly $f _ n \rightrightarrows _ {\Omega\backslash E _ \epsilon} f$.

Convergence in Measure #

Convergence in measure locally #

Suppose $(\Omega, \A, \mu)$ is a $\sigma$-finite measure space, $E$ is a metric space. Let $f, (f _ n) _ {n = 1}^\infty \in \L(\Omega \to E)$.

• (Fast I) Suppose $\forall \epsilon > 0, \forall A \in \A^<:\sum _ {n = 1}^\infty \mu _ A(\{d(f, f _ n) > \epsilon\}) < \infty$. Then $f _ n \to f$ $\mu$-a.e.
  • Suppose $A \in \A^<$. Let $B _ {n, \epsilon} = A \cap \{d(f, f _ n) > \epsilon\}$ and $B _ {\epsilon} = \limsup _ n B _ {n,\epsilon}$.
  • Then $\mu _ A(B _ \epsilon) = 0$ by (Cantelli).
  • Define $N = \cup _ n B _ {1/n}$, $\mu(N) = 0$, and $f _ n \to f$ on $A \backslash N$.
• Since $\mu$ is $\sigma$-finite, repeat above process on each patch of $\Omega$.
• (Fast II) Suppose $E$ is complete. Suppose $(\epsilon _ n) _ {n = 1}^\infty > 0$, $\sum _ n \epsilon _ n < \infty$. And $\forall A \in \A^<: \sum _ {n = 1}^\infty \mu _ A\{d(f _ n, f _ {n + 1}) > \epsilon _ n\} <\infty$. Then $f _ n \to f$ $\mu$-a.e. for some $f \in \L(\Omega \to E)$.
  • Now prove in $(A, \A| _ A, \mu _ A)$.
  • Let $B _ n = \{d(f _ n, f _ {n + 1}) > \epsilon _ n\}$ and $B = \limsup _ n B _ n$.
  • Since $\sum _ n \mu B _ n < \infty$, $\mu _ A(B) = 0$ by (Cantelli).
  • For $\omega \in A \backslash B$, $d(f _ n, f _ {n + 1}) > \epsilon _ n$ finitely many times. Now $f _ n(\omega)$ is Cauchy in $E$ on $A \backslash B$.
  • So for $f _ n \to f$ $\mu$-a.e. on $A$.
• (Metric of convergence) There is a metric $\tilde d$ on $\L(\Omega \to E)$, s.t. $f _ n \to f$ in $\mu$ iff $\tilde d(f, f _ n) \to 0$.
  • Suppose $(A _ N) _ {N = 1}^\infty \in \A^<$ where $A _ N\uparrow \Omega$.
  • Define $\tilde d _ N(f,g) = \mu\min(1 _ {A _ N}, d(f, g))$.
  • Define $\tilde d(f, g) = \sum _ N \tilde d _ N (f, g)2 ^{-N}/(1 + \mu A _ N)$.
  • Now $0 \le \tilde d _ N(f, g) \le \mu A _ N$ and $0\le \tilde d(f, g) \le 1$.
  • It is easy to verify that $\tilde d$ is indeed a metric on $\L(\Omega \to E, \A)$.
  • For $f, (f _ n) _ {n = 1}^\infty \in \L(\Omega \to E, \A)$, $f _ n \to f$ in $\mu$ if and only if $\tilde d(f, f _ n) \to 0$.
    • $\tilde d(f, f _ n) \to 0$ if and only if $\forall N \ge 1: \tilde d _ N(f, f _ n) \to 0$.
    • $\tilde d _ N(f, f _ n) \to 0$ if and only if $\forall \epsilon > 0:\mu(\{d(f, f _ n) > \epsilon\} \cap A _ N) \to 0$.
• (Subsequence) $f _ n \to f$ in $\mu$ if and only if for any subsequence $(f _ {k})$ of $(f _ n)$, there is a subsequence $(f _ m)$ such that $f _ m \to f$ $\mu$-a.e.
  • $\to$ direction: Pick $(f _ l) \subset (f _ m)$ such that $\mu(A _ l \cap \{d(f, f _ l) > 1/l\}) < 2^{-l}$.(Fast I).
  • $\leftarrow$ direction: Prove the contrapositive. Conside the metric space $(\L(\Omega \to E), \tilde d)$.
• (Cauchy) $(f _ n)$ is Cauchy in $\mu$ if $\forall A \in \A^<,\forall \epsilon > 0: \lim _ {m, n \to \infty}\mu _ A\{d(f _ m, f _ n) > \epsilon\} = 0$.
• (Completeness, TODO) $(f _ n)$ is Cauchy in $\mu$ then $f _ n \to f$ in $\mu$ for some $f \in \L(\Omega \to E)$.

Convergence in measure globally #

Let $(E, d)$ be a metric sapce. And $(\Omega, \A, \mu)$ is a measure space. Suppose $f, (f _ n) _ {n = 1}^\infty \in \L(\Omega \to E)$.

• (Cauchy) $(f _ n)$ is Cauchy in $\mu$ globally if $\forall \epsilon > 0: \lim _ {m, n \to \infty} \mu\{d(f _ n, f _ m) > \epsilon\} = 0$.
  • If $f _ n \to f$ in $\mu$ then $(f _ n)$ is Cauchy in $\mu$.
• (Subsequence) TODO $f _ n \to f$ in $\mu$ globally, then there exists $(f _ k) \subset (f _ n)$ where $f _ k \rightrightarrows f$ in $\mu$.
  • See Ash. pp.97 for the proof.

Product and convergence in measure TODO #

See this stackexchange page.

Suppose $f, (f _ n) _ {n = 1}^\infty, g, (g _ n) _ {n = 1}^\infty \in \L(\Omega \to \R, \mu)$. Where $f _ n \to f$ and $g _ n \to g$ in $\mu$, and $\mu$ is a finite measure.

Then $f _ n g _ n \to fg$ in measure.

Convergence in Norm #

Convergence in norm #

Suppose $(\Omega, \A, \mu)$ is a measure space. Suppose $f, (f _ n) _ {n = 1}^\infty \in \L^p(\Omega \to \R, \mu)$ for $p \in (0, \infty)$.

• (Fast III) Suppose $\mu$ is $\sigma$-finite. $\sum _ {n = 1}^\infty \|f _ n - f\|^p _ p < \infty$. Then $f _ n \to f$ $\mu$-a.e.
  • By (Markov) $\mu(\{|f-f _ {n}|>\varepsilon\}) \leq \varepsilon^{-p}\|f-f _ {n}\| _ {p}^{p}$. Now apply (Fast I).

Suppose $p \in [1, \infty)$. $L^p$ is a Banach space.

• (Cauchy) $(f _ n)$ is called Cauchy in $\L _ p$ if $\lim _ {m, n \to \infty} \|f _ n - f _ m\| _ p = 0$.
  • $(f _ n)$ is Cauchy in $\L _ p$ then $(f _ n)$ is Cauchy in $\mu$. (Hint: by converse).
  • $f _ n \to f$ in $\L _ p$ then $(f _ n)$ is Cauchy in $\L _ p$.
• (Fast IV) Suppose $\|f _ k - f _ {k + 1}\| _ p < 4^{-k}$. Then $f _ n \to f$ $\mu$-a.e. to some $f \in \L(\Omega \to \R)$.
  • Let $A _ k = \{|f _ k - f _ {k + 1}| \ge 2^{-k}\}$. By (Markov) $\mu A _ k \le 2^{-kp}$.
  • By (Cantelli) $\mu A^ * = 0$, so $f _ k(\omega)$ is cauchy $\mu$-a.e.
  • So $f _ n$ converges to some $f \in \L(\Omega \to \R)$.
• (Completeness) $(f _ n)$ is Cauchy in $\L _ p$ then $f _ n \to f$ in $\L _ p$ norm to some $f \in \L^p(\Omega \to \R)$.
  • Pick $(f _ {n _ k}) \subset (f _ n)$ according to (Fast IV), then $f _ {n _ k} \to f$ $\mu$-a.e. for some $f \in \L(\Omega \to \R)$.
  • $\forall \epsilon > 0, \exists N \in \N: \forall n, m \ge N: \|f _ n - f _ m\| _ p^p < \epsilon$.
  • $\liminf _ {k \to \infty} \|f _ n - f _ {n _ k}\| _ p^p = \liminf _ {k \to \infty} \mu |f _ n - f _ {n _ k}|^p \ge \mu \liminf _ {k \to \infty} \mu|f _ n -f|^p = \|f _ n - f\| _ p^p$
  • So $\|f _ n - f\| _ p \to 0$. So $f \in \L^p(\Omega \to \R)$.

Suppose $p = \infty$. $\L^\infty$ is a Banach space.

• (Completeness) $\L^\infty$ is a Banach space.
  • Since $\bF$ is complete, Cauchy essentially pointwise implies essentially uniform convergence.
  • Essentially uniform convergence if and only if convergence in $\L^\infty$.
• (Completeness) $(f _ n)$ is Cauchy in $\L _ \infty$ iff $f _ n \to f$ in $\L _ \infty$ norm for some $f \in \L^\infty (\Omega \to \R)$.
  • Since $\|f _ n - f _ m\| _ \infty \to 0$ as $n, m \to \infty$.
  • $\forall \epsilon > 0, \exists N \in \N, \forall n, m > N, \forall x \in A: |f _ n(x) - f _ m(x)| \le \epsilon$. Where $\mu(A^c) = 0$.
  • So $f _ n \to f$ $\mu$-a.e. for some $f \in \L(\Omega \to \R)$.
  • The convergence is uniform on $A$ since $\lim _ {m \to \infty} |f _ n(x) - f(x)| \le \epsilon$.
  • So $f \in \L^\infty(\Omega \to \R)$.

Uniqueness of Convergence #

Implication diagram #

Notice that L-finite and L-infinity are only defined for codomain $\symbf F$ is $\R$ or $\C$.

graph
A(Pointwise)
B(Uniform)
C(Essentially pointwise)
D(Essentially uniform)
E(Almost uniform)
F(Locally in measure)
G(Globally in measure)
H{{L-finite}}
I{{L-infinity}}
B --> A
A --> C
B --> D
D --> C
B --> E
D --> I
I --> D
D --> E
H --> G
G --> F
E --> G
C --> F
E --> C
C -.-> E
F -.-> C
G -.->E

Convergence in $\symbf F$ #

Suppose $(\Omega, \A, \mu)$ is a measure space. And $\symbf F = \R, \C$, $c \in \bF$.

Suppose $(f _ n) _ {n = 1}^\infty, (g _ n) _ {n = 1}^\infty, f, g \in \L(\Omega \to \symbf F)$. All convergence modes are well defined. Then

• $f _ n \to f$ in any mode if and only if $|f _ n - f| \to 0$ in any mode.
• $f _ n \to f$ and $g _ n \to g$ in some mode then $f _ n + g _ n \to f + g$ and $cf _ n \to cf$ in the same mode.
• $f _ n \to 0$ in some mode and $\forall n \in \N^+: |g _ n| \le f _ n$ then $g _ n \to 0$ in the same mode.

Uniqueness of convergence #

Suppose $(\Omega, \A, \mu)$ is a measure space and $(E, d)$ is a metric space.

Suppose $(f _ n) _ {n = 1}^\infty, (g _ n) _ {n = 1}^\infty, f, g \in \L(\Omega \to E, \A/\B(E))$.

• When $\mu$ is $\sigma$-finite. $f _ n \to f$ in $\mu$ locally and $f _ n \to g$ in $\mu$ locally implies $f = g$ $\mu$.a.e.
  • $\mu _ A\{d(f, g) > \epsilon \} \le \mu _ A\{d(f, f _ n) > \epsilon / 2\} + \mu _ A\{d(g, f _ n) > \epsilon / 2\} \to 0$.
  • So $\mu(\{d(f, g) > 0\}) = 0$ by union of each region.
• $f _ n \to f$ in $\mu$ globally and $f _ n \to g$ gloablly implies $f = g$ $\mu$.a.e.
  • The proof is similar to the above one.
• $f _ n \to f$ $\mu$.a.e. and $f _ n \to g$ $\mu$.a.e. implies $f = g$ $\mu$.a.e. (Apparently.)
• $f _ n \to g$ in $\mu$ globally and $f _ n \to f$ $\mu$.a.e. implies $f = g$ $\mu$.a.e.
  • Prove by contradiction, suppose $A = \{d(f, g) \neq 0\}$ is not a null set.
  • Exists $\epsilon > 0$ where $\mu\{d(f, g) > \epsilon\} = c > 0$.
  • $\mu\{d(f _ n, g) > \epsilon\} \to 0$. But since $d(f _ n, g) + d(f _ n, f) \ge d(f, g)$.
  • So $\mu\{d(f, g) > \epsilon\} \le \mu\{d(f _ n, g) > \epsilon / 2\} + \mu\{d(f _ n, f) > \epsilon / 2\} \to 0$. Contradiction!

Corollary of DCT #

Suppose $(f _ n) _ {n = 1}^\infty \in \L^p(X \to \eR, \mu)$ where $p \in (0, \infty]$, and $f \in \L^p(X \to \eR)$.

Suppose $f _ n \to f$ $\mu$-a.e. and for $g \in \L^p(X \to [0, \infty], \mu)$, $|f _ n| \le g$. Then $\mu |f _ n - f|^p \to 0$.

• $|f _ n - f|^p \le (|f _ n| + |f|)^p \le (2|g|)^p$. And $|f _ n - f|^p \downarrow 0$.
• By (DCT) $\mu |f _ n - f|^p \to 0$ (or $f _ n \to f$ in $\L^p$).