Prove the Lebesgue integral is translation/dilation invariant.
Prove continuity in \(L_1\): \(\norm{\tau_hf - f}\converges{h\to 0}\too 0\).
Prove that \(E\) is measurable \(\iff\) \(E = F \disjoint Z\) with \(F\in F_\sigma\) and \(Z\) null \(\iff\) \(E = G\sm Z\) with \(G\in G_\delta\) and \(Z\) null.
Show that \(m(E) = \sup_{K \subseteq E}m(K) \iff\) there exists \(K = K(\eps)\) with \(m(K) \in [m(E) - \eps, m(E)]\).
- What’s most useful here is the proof technique, not so much the result itself.
Apply Fubini and Tonelli to literally anything.
Prove that \(\norm{f}_p\to \norm{f}_\infty\) over a finite measure space.
Apply Cauchy-Schwarz to literally anything, in the form of \(\norm{fg}_1 \leq \norm{f}_2 \norm{g}_2\).
Solution
Let \(m\) be Lebesgue measure on \(\RR^n\), \(\tau_h f(x) = f(x + h)\), and \(f_\delta(x) = \delta^{-n}f(x/\delta)\) for \(\delta > 0\).
1. For measurable \(f \geq 0\) or \(f \in L^1\), \(\int \tau_h f = \int f\) and \(\int f_\delta = \int f\).
1.1. The identities hold for \(f = \chi_E\) with \(E\) measurable.
\(\tau_h\chi_E = \chi_{E-h}\) and \((\chi_E)_\delta = \delta^{-n}\chi_{\delta E}\), and Lebesgue measure satisfies \(m(E - h) = m(E)\) and \(m(\delta E) = \delta^n m(E)\).
1.2. Q.E.D.
By linearity, step 1 gives the identities for nonnegative simple functions. For measurable \(f \geq 0\) choose simple \(0 \leq s_k \nearrow f\); then \(\tau_h s_k \nearrow \tau_h f\) and \((s_k)_\delta \nearrow f_\delta\), and the monotone convergence theorem passes the identities to \(f\). For \(f \in L^1\) apply this to \(f^+\) and \(f^-\).
2. For \(f \in L^1\), \(\|\tau_h f - f\|_1 \to 0\) as \(h \to 0\).
2.1. The claim holds for \(f = \chi_I\) with \(I\) a bounded box.
\(\|\tau_h\chi_I - \chi_I\|_1 = m(I \triangle (I - h))\), and \(I \triangle (I-h)\) is contained in the set of points within distance \(|h|\) of \(\partial I\), whose measure tends to \(0\) with \(h\).
2.2. Step functions, finite linear combinations of indicators of bounded boxes, are dense in \(L^1\).
Simple functions are dense in \(L^1\). For \(m(E) < \infty\) and \(\eps > 0\), outer regularity gives an open \(U \supseteq E\) with \(m(U \setminus E) < \eps\). \(U\) is a countable union of almost disjoint closed cubes, so a finite union \(A\) of them has \(m(U \setminus A) < \eps\), and \(\|\chi_E - \chi_A\|_1 < 2\eps\).
2.3. Q.E.D.
Given \(\eps > 0\), step 2 gives a step function \(s\) with \(\|f - s\|_1 < \eps/3\), and step 1 with the triangle inequality gives \(\|\tau_h s - s\|_1 < \eps/3\) for \(|h|\) small. By step 1, \(\|\tau_h f - \tau_h s\|_1 = \|f - s\|_1\), so \(\|\tau_h f - f\|_1 < \eps\) for \(|h|\) small.
3. \(E\) is measurable if and only if \(E = F \sqcup Z\) with \(F\) an \(F_\sigma\) set and \(Z\) null, if and only if \(E = G \setminus Z'\) with \(G\) a \(G_\delta\) set and \(Z'\) null.
3.1. If \(E\) is measurable, then \(E = F \sqcup Z\) as stated.
Outer regularity applied to \(E^c\) gives closed \(F_k \subseteq E\) with \(m(E \setminus F_k) < 1/k\). Put \(F = \bigcup_k F_k\) and \(Z = E \setminus F\). Then \(F\) is \(F_\sigma\), \(F \cap Z = \emptyset\), and \(m(Z) \leq m(E \setminus F_k) < 1/k\) for every \(k\).
3.2. If \(E\) is measurable, then \(E = G \setminus Z'\) as stated.
Outer regularity gives open \(G_k \supseteq E\) with \(m(G_k \setminus E) < 1/k\). Put \(G = \bigcap_k G_k\) and \(Z' = G \setminus E\). Then \(G\) is \(G_\delta\), \(E = G \setminus Z'\), and \(m(Z') \leq m(G_k \setminus E) < 1/k\) for every \(k\).
3.3. Q.E.D.
Steps 1 and 2 are the forward implications. Conversely, \(F_\sigma\) and \(G_\delta\) sets are Borel, null sets are Lebesgue measurable, and unions and differences of measurable sets are measurable.
4. \(m(E) = \sup\theset{m(K) : K \subseteq E \text{ compact}}\) if and only if for every \(\eps > 0\) there is a compact \(K \subseteq E\) with \(m(E) - \eps \leq m(K) \leq m(E)\).
Every compact \(K \subseteq E\) has \(m(K) \leq m(E)\), so \(m(E)\) is an upper bound of the set \(S = \theset{m(K)}\). The supremum equals the upper bound \(m(E)\) exactly when no \(m(E) - \eps\) with \(\eps > 0\) is an upper bound of \(S\), which is the stated condition.
5. If \(\mu(X) < \infty\) and \(f \in L^\infty(\mu)\), then \(\|f\|_p \to \|f\|_\infty\) as \(p \to \infty\).
If \(\mu(X) = 0\) both sides vanish; assume \(\mu(X) > 0\). Upper bound: \(\|f\|_p \le \|f\|_\infty\mu(X)^{1/p} \to \|f\|_\infty\). Lower bound: for \(0 < M < \|f\|_\infty\) the set \(\theset{|f| > M}\) has positive finite measure, and \(\|f\|_p \ge M\mu\theset{|f| > M}^{1/p} \to M\). So \(M \le \liminf_p \|f\|_p \le \limsup_p \|f\|_p \le \|f\|_\infty\) for every such \(M\).