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/Part 3: Functional Analysis/Chapter 15: $L^{p}$ Spaces

15.4 Uniform Integrability

Definition 15.4.1 (Uniform Integrability).label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\cf \subset L^{p}(X; E)$, then $\cf$ is uniformly $p$-integrable if

\[\lim_{M \to \infty}\sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}^{p} d\mu = 0\]

Proposition 15.4.2.label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\cf \subset L^{p}(X; K)$, and let $|\cf|^{p} = \bracsn{\norm{f}^p| f \in \cf}$, then:

  1. (1)

    If $\cf$ is uniformly $p$-integrable, then $|\cf|^{p}$ is uniformly absolutely continuous with respect to $\mu$.

  2. (2)

    If $\sup_{f \in \cf}\norm{f}_{L^p(X; E)}< \infty$ and $|\cf|^{p}$ is uniformly absolutely continuous with respect to $\mu$, then $\cf$ is uniformly $p$-integrable.

Proof. (1): Let $\eps > 0$, then there exists $M \ge 0$ such that

\[\sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}^{p} d\mu < \eps/2\]

Thus for any $A \in \cm$,

\begin{align*}\sup_{f \in \cf}\int_{A}\norm{f}^{p} d\mu&\le \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \le M} \cap A}\norm{f}^{p} d\mu+ \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M} \cap A}\norm{f}^{p} d\mu \\&\le M \mu(A) + \eps/2\end{align*}

so if $\mu(A) \le M\eps/2$, then $\sup_{f \in \cf}\int_{A}\norm{f}^{p} d\mu \le \eps$.

(2): Let $\eps > 0$, then there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\sup_{f \in \cf}\int_{A} \norm{f}^{p} d\mu < \eps$. By Markov’s inequality,

\[\sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) \le \frac{\sup_{f \in \cf}\norm{f}_{L^p(X; E)}^{p}}{M^{p}}\]

Thus for sufficiently large $M$, $\sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) < \delta$, and

\[\sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}^{p} d\mu < \eps\]

$\square$

Theorem 15.4.3 (Vitali Convergence Theorem).label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^{p}(X; E)$ if and only if:

  1. (M)

    $\fF$ is locally Cauchy in measure.

  2. (UI)

    For each $\eps > 0$, there exists $M \ge 0$ and $F \in \fF$ such that

    \[\sup_{f \in F}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_{E}^{p} d\mu < \eps\]

  3. (T)

    For each $\eps > 0$, there exists $A \in \cm$ and $F \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F}\int_{A^c}\norm{f}_{E}^{p} < \eps$.

Proof. ($L^{p}$) $\Rightarrow$ (M): By Markov’s inequality.

($L^{p}$) $\Rightarrow$ (UI): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)}< \eps$. Fix $f \in F$, then by the Dominated Convergence Theorem, there exists $M > 0$ such that $\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_{E}^{p} d\mu < \eps$. For any $g \in F$,

\[\bracs{\norm{g}_E \ge 2M}\subset \bracs{\norm{f}_E \ge M}\cup \bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}\]

so by Lemma 15.2.2,

\begin{align*}\int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_{E}^{p}d\mu&\le \int_{\bracs{\norm{f}_E \ge M}}\norm{g}_{E}^{p}d\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{g}_{E}^{p}d\mu \\&\le \int_{\bracs{\norm{f}_E \ge M}}\norm{f}_{E}^{p}d\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_{E}^{p}d\mu \\&+ 2p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)}\vee \norm{g}_{L^p(X; E)})^{p-1}\end{align*}

Since $\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}\subset \bracs{\norm{f - g}_E \ge M}$, by Markov’s inequality,

\[\int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_{E}^{p}d\mu \le M^{p}\mu\bracs{\norm{f - g}_E \ge M}\le \norm{f - g}_{L^p(X; E)}^{p}\]

Therefore

\[\sup_{g \in F}\int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_{E}^{p} d\mu \le 2p \eps (\norm{f}_{L^p(X; E)}+ \eps)^{p - 1}+ \eps + \eps^{p}\]

($L^{p}$) $\Rightarrow$ (T): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)}< \eps$. Fix $f \in F$, then by the Dominated Convergence Theorem, there exists $A \in \cm$ such that $\mu(A) < \infty$ and $\norm{\one_{A^c}f}_{L^p(X; E)}< \eps$. In which case, for any $g \in F$,

\[\norm{\one_{A^c}g}_{L^p(X; E)}\le \norm{\one_{A^c}f}_{L^p(X; E)}+ \norm{f - g}_{L^p(X; E)}\le \norm{\one_{A^c}f}_{L^p(X; E)}+ \eps\]

(M) + (UI) + (T) $\Rightarrow$ ($L^{p}$): Let $\eps > 0$. By (T), there exists $A \in \cm$ and $F_{1} \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F_1}\int_{A^c}\norm{f}_{E}^{p} < \eps^{p}$. Thus for every $f, g \in F_{1}$,

\begin{align*}\norm{f - g}_{L^p(X; E)}&\le \norm{\one_A(f - g)}_{L^p(X; E)}+ \norm{\one_{A^c}f}_{L^p(X; E)}+ \norm{\one_{A^c}g}_{L^p(X; E)}\\&\le \norm{\one_A(f - g)}_{L^p(X; E)}+ 2\eps\end{align*}

By (UI), there exists $M > 0$ and $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that

\[\sup_{h \in F_2}\int_{\bracs{\norm{h}_E \ge M}}\norm{h}_{E}^{p} < \eps^{p}\]

Assume without loss of generality that $\mu(A) > 0$ and let $\delta = \eps\mu(A)^{-1/p}$. By (M), there exists $F_{3} \in \fF$ with $F_{3} \subset F_{2}$, such that for any $f, g \in F_{3}$,

\[\mu(A \cap \bracsn{\norm{f - g}_E \ge \delta}) \le \paren{\frac{\eps}{2M}}^{p}\]

In which case,

\begin{align*}\norm{\one_{A}(f - g)}_{L^p(X; E)}&\le \normn{\one_{A \cap \bracs{\norm{f - g}_E \le \delta}}(f - g)}_{L^p(X; E)}\\&+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}\\&\le \delta\mu(A)^{1/p}+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}\\&\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}+ \eps\end{align*}

then for any $f, g \in F_{3}$,

\begin{align*}\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}&\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)}\\&+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)}\end{align*}

Now,

\begin{align*}\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)}&\le \normn{\one_{\bracsn{\norm{f}_E \ge M}}f}_{L^p(X; E)}\\&+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta, \norm{f}_E \le M}}f}_{L^p(X; E)}\\&\le \eps + \eps/2 = 3\eps/2\end{align*}

Similarly, $\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)}\le 3\eps/2$. Thus

\[\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}\le 3\eps\]

Therefore for any $f, g \in F_{3}$, $\norm{f - g}_{L^p(X; E)}\le 6 \eps$, so $\fF$ is Cauchy in $L^{p}(X; E)$.$\square$

Corollary 15.4.4 (Dominated Convergence Theorem (In Measure)).label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^{p}(X; \real)$ such that:

  1. (M)

    $\fF \to g$ locally in measure.

  2. (D)

    There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$.

then $\fF \to f$ in $L^{p}(X; E)$. In particular, if $p = 1$, then

\[\lim_{f, \fF}\int f d\mu = \int g d\mu\]

Proof. Since (D) implies (UI) and (T) of the Vitali Convergence Theorem, the result follows from the Vitali Convergence Theorem.$\square$

Lemma 15.4.5 (Scheffé).label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^{p}(X; E)$, then $\fF \to g$ in $L^{p}(X; E)$ if and only if:

  1. (M)

    $\fF \to g$ locally in measure.

  2. (N)

    $\lim_{f, \fF}\norm{f}_{L^p(X; E)}= \norm{g}_{L^p(X; E)}$.

Proof. It is sufficient to show conditions (UI) and (T) of the Vitali Convergence Theorem.

(T): Let $\eps > 0$. By (N), there exists $F_{1} \in \fF$ such that $|\norm{f}_{L^p(X; E)}^{p} - \norm{g}_{L^p(X; E)}^{p}| < \eps$ for all $f \in F_{1}$. By the Dominated Convergence Theorem, there exists $A \in \cm$ with $\mu(A) < \infty$ such that:

  1. (i)

    $\int_{A^c}\norm{g}_{E}^{p} < \eps$.

Since $\norm{g}_{E}^{p} d\mu \ll \mu$, by Proposition 22.4.3, there exists $\delta > 0$ such that:

  1. (ii)

    For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_{E}^{p} d\mu < \eps$.

  2. (iii)

    $\int_{A}[(\norm{g}_{E} - \delta) \vee 0]^{p} d\mu > \int_{A} \norm{g}_{E}^{p} d\mu - \eps$.

By (M), there exists $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that for each $f \in F_{2}$, $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. In which case, for any $f \in F_{2}$,

\begin{align*}\int_{A} \norm{f}_{E}^{p} d\mu&\ge \int_{A \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_{E}^{p} d\mu \\&\ge \int_{A}[(\norm{g}_{E} - \delta) \vee 0]^{p} d\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_{E}^{p} d\mu\end{align*}

By (iii),

\[\int_{A} \norm{f}_{E}^{p} d\mu \ge \int_{A} \norm{g}_{E}^{p} d\mu - \eps - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_{E}^{p} d\mu\]

and by (ii), $\int_{A} \norm{f}_{E}^{p} d\mu \ge \int_{A}\norm{g}_{E}^{p} d\mu - 2\eps$. Finally, by (i),

\[\int_{A} \norm{f}_{E}^{p} d\mu \ge \int \norm{g}_{E}^{p} d\mu - 3\eps \ge \int \norm{f}_{E}^{p} d\mu - 4\eps\]

and $\int_{A^c}\norm{f}_{E}^{p} d\mu \le 4\eps$.

(UI): Let $\eps > 0$. By (N) and (T), there exists $F_{1} \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that for every $f \in F_{1}$,

  1. (i)

    $|\norm{f}_{L^p(X; E)}^{p} - \norm{g}_{L^p(X; E)}^{p}| < \eps$.

  2. (ii)

    $\int_{A^c}\norm{f}_{E}^{p} d\mu, \int_{A^c}\norm{g}_{E}^{p} d\mu < \eps$.

By (i) and (ii),

  1. (iii)

    $\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p}d\mu \le 3\eps$.

Since $\norm{g}_{E}^{p} d\mu \ll \mu$, by Proposition 22.4.3, there exists $\delta > 0$ such that:

  1. (iv)

    For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_{E}^{p} d\mu < \eps$.

  2. (v)

    $\int_{A}[(\norm{g}_{E} - \delta) \vee 0]^{p} d\mu > \int_{A} \norm{g}_{E}^{p} d\mu - \eps$.

By (M), there exists $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that for every $f \in F_{2}$,

  1. (vi)

    $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$.

Let $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$, then for any $f \in F_{2}$,

\begin{align*}\int_{A \setminus B}\norm{f}_{E}^{p} d\mu&\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_{E}^{p} d\mu \\&\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}(\norm{g}_{E} - \delta \vee 0)^{p}d\mu \\&\ge \int_{(A \setminus B)}(\norm{g}_{E} - \delta \vee 0)^{p}d\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_{E}^{p} d\mu\end{align*}

By (vi) and (iv), $\int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_{E}^{p} d\mu < \eps$, so

\[\int_{A \setminus B}\norm{f}_{E}^{p} d\mu \ge \int_{(A \setminus B)}(\norm{g}_{E} - \delta \vee 0)^{p}d\mu - \eps\]

By (v),

\[\int_{A \setminus B}\norm{f}_{E}^{p} d\mu \ge \int_{(A \setminus B)}\norm{g}_{E}^{p}d\mu - 2\eps\]

Since $\mu(B) < \delta$, by (iv),

\[\int_{A \setminus B}\norm{f}_{E}^{p} d\mu \ge \int_{A}\norm{g}_{E}^{p}d\mu - 3\eps\]

and by (iii),

\[\int_{A \setminus B}\norm{f}_{E}^{p} d\mu \ge \int_{A}\norm{f}_{E}^{p}d\mu - 6\eps\]

so $\int_{B}\norm{f}_{E}^{p} d\mu \le 6\eps$ for all $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$.

Finally, by (i) and Markov’s Inequality, there exists $M \ge 0$ such that $\mu\bracs{\norm{f}_E \ge M}< \delta$ for all $f \in F_{2}$. Therefore for any $f \in F_{2}$,

\[\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_{E}^{p} d\mu \le \int_{A \cap \bracs{\norm{f}_E \ge M}}\norm{f}_{E}^{p} d\mu + \int_{A^c}\norm{f}_{E}^{p} d\mu \le 7\eps\]

$\square$

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