Show that continuity of measure from above/below holds for outer measures.
Show that a countable union of null sets is null.
Measurability
- Show that \(f=0\) a.e. iff \(\int_E f = 0\) for every measurable set \(E\).
Integrability
Show that if \(f\) is a measurable function, then \(f=0\) a.e. iff \(\int f = 0\).
Show that a bounded function is Lebesgue integrable iff it is measurable.
Show that simple functions are dense in \(L^1\).
Show that step functions are dense in \(L^1\).
Show that smooth compactly supported functions are dense in \(L^1\).
Erratum: continuity from below holds for Lebesgue outer measure (step <1>1), but not for every outer measure; see E-ASWCD. Continuity from above fails even for Lebesgue outer measure on sets of finite outer measure (step <1>2); it holds for measurable sets. The integral statements need \(f \ge 0\): for \(f = \chi_{[0,1]} - \chi_{[1,2]}\), \(\int f = 0\) but \(f \neq 0\) on a set of measure \(2\). The boundedness statement needs a finite measure space.
Solution
Let \(m^*\) be Lebesgue outer measure on \(\RR^n\) and \(m\) Lebesgue measure. For every \(A \subseteq \RR^n\) and \(\eps > 0\) there is a measurable \(B \supseteq A\) with \(m(B) \le m^*(A) + \eps\), by outer regularity.
1. If \(E_1 \subseteq E_2 \subseteq \cdots\) and \(E = \bigcup_j E_j\), then \(m^*(E_j) \to m^*(E)\).
Monotonicity gives \(m^*(E_j) \le m^*(E_{j+1}) \le m^*(E)\). Fix \(\eps > 0\) and choose measurable \(B_k \supseteq E_k\) with \(m(B_k) \le m^*(E_k) + \eps\). Put \(C_j = \bigcap_{k \ge j} B_k\). Since \(E_j \subseteq E_k \subseteq B_k\) for \(k \ge j\), \(C_j \supseteq E_j\); the \(C_j\) increase; and \(m(C_j) \le m(B_j)\). So \(E \subseteq \bigcup_j C_j\), and continuity from below for \(m\) gives \(m^*(E) \le \lim_j m(C_j) \le \lim_j m^*(E_j) + \eps\).
2. Continuity from above fails for \(m^*\): there are \(E_1 \supseteq E_2 \supseteq \cdots\) in \([-1, 2]\) with \(\bigcap_j E_j = \emptyset\) and \(\inf_j m^*(E_j) > 0\).
Let \(N \subseteq [0,1]\) contain exactly one point of each coset of \(\QQ\) in \(\RR\) meeting \([0,1]\), a Vitali set; \(N\) is not measurable, so \(m^*(N) > 0\). Let \((q_k)\) enumerate \(\QQ \cap [-1,1]\); the translates \(N + q_k\) are pairwise disjoint. Put \(E_j = \bigcup_{k \ge j}(N + q_k)\). Then \(E_j\) decreases, a point of \(\bigcap_j E_j\) would lie in \(N + q_k\) for infinitely many \(k\), so \(\bigcap_j E_j = \emptyset\), and \(m^*(E_j) \ge m^*(N + q_j) = m^*(N)\).
3. A countable union of null sets is null.
\(m^*(\bigcup_k E_k) \le \sum_k m^*(E_k) = 0\) by countable subadditivity.
4. For measurable \(f \ge 0\), \(f = 0\) a.e. if and only if \(\int f = 0\), if and only if \(\int_E f = 0\) for every measurable \(E\).
If \(f = 0\) a.e., then \(f\chi_E = 0\) a.e. and \(\int_E f = 0\) for every \(E\), in particular for \(E = \RR^n\). If \(\int f = 0\), then \(\int f \ge \frac1n m\theset{f \ge 1/n}\) gives \(m\theset{f \ge 1/n} = 0\) for every \(n\), and \(\theset{f > 0} = \bigcup_n \theset{f \ge 1/n}\) is null.
5. On a measure space with \(\mu(X) < \infty\), a bounded function is integrable if and only if it is measurable.
Lebesgue integrability is defined for measurable functions. If \(f\) is measurable and \(|f| \le M\), then \(\int |f| \le M\mu(X) < \infty\).
6. Simple functions are dense in \(L^1\).
It suffices to approximate \(g \in L^1\) with \(g \ge 0\), and then \(f^+\) and \(f^-\) separately. The simple functions \(s_n = \min\big(2^{-n}\lfloor 2^n g\rfloor, n\big)\) satisfy \(0 \le s_n \le g\) and \(s_n \to g\) wherever \(g < \infty\), hence a.e., so dominated convergence with dominating function \(g\) gives \(\|g - s_n\|_1 \to 0\).
7. Step functions are dense in \(L^1(\RR)\).
By step 6 it suffices to approximate \(\chi_E\) with \(m(E) < \infty\). Outer regularity gives an open \(G \supseteq E\) with \(m(G \setminus E) < \eps\). \(G\) is a countable disjoint union of open intervals \(I_k\), and \(G_N = \bigcup_{k \le N} I_k\) satisfies \(m(G \setminus G_N) \to 0\). So \(\|\chi_{G_N} - \chi_E\|_1 = m(G_N \triangle E) < 2\eps\) for large \(N\).
8. \(C_c^\infty(\RR^n)\) is dense in \(L^1(\RR^n)\).
Given \(f \in L^1\), dominated convergence gives \(R\) with \(\|f - f\chi_{B(0,R)}\|_1 < \eps/2\). Let \(\phi \ge 0\) be smooth with compact support and \(\int \phi = 1\), and \(\phi_\delta(x) = \delta^{-n}\phi(x/\delta)\). Then \((f\chi_{B(0,R)}) \ast \phi_\delta\) is smooth with compact support, and \(\|(f\chi_{B(0,R)}) \ast \phi_\delta - f\chi_{B(0,R)}\|_1 \to 0\) as \(\delta \to 0\) by continuity of translation in \(L^1\).