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: the first part needs \(f \geq 0\) (or \(|f|\) in place of \(f\)). For \(f = \chi_{[0,1]} - \chi_{[1,2]}\), \(\int f = 0\) but \(f \neq 0\) on a set of measure \(2\). The second part needs a finite measure space: on \(\RR\) with Lebesgue measure, \(f \equiv 1\) is bounded and measurable but not integrable.
Solution
1. For measurable \(f \geq 0\), \(f = 0\) a.e. if and only if \(\int f = 0\).
If \(f = 0\) a.e., every simple function \(0 \leq s \leq f\) vanishes a.e. and has integral \(0\), so \(\int f = 0\). Conversely, \(\theset{f > 0} = \bigcup_{k} \theset{f > 1/k}\), and \(\int f \ge \frac{1}{k}\mu\theset{f > 1/k}\) for each \(k\); if \(\int f = 0\), each \(\theset{f > 1/k}\) is null, hence so is \(\theset{f > 0}\).
2. On a measure space \((X,\mu)\) with \(\mu(X) < \infty\), a bounded function is Lebesgue integrable if and only if it is measurable.
The Lebesgue integral is defined for measurable functions, so an integrable function is measurable. Conversely, if \(f\) is measurable and \(|f| \le M\), then \(\int|f| \le M\mu(X) < \infty\).
3. Simple functions are dense in \(L^1\).
For \(f \geq 0\) in \(L^1\), the dyadic simple functions \[ s_k = \sum_{j=1}^{k 2^k}\frac{j-1}{2^k}\chi_{\theset{\frac{j-1}{2^k} \le f < \frac{j}{2^k}}} + k\chi_{\theset{f \ge k}} \] satisfy \(0 \le s_k \nearrow f\) pointwise, so \(|f - s_k| \le f\) and the dominated convergence theorem gives \(\|f - s_k\|_1 \to 0\). For general \(f\), approximate \(f^+\) and \(f^-\) separately.
4. Step functions, finite linear combinations of indicators of bounded intervals, are dense in \(L^1(\RR)\).
4.1. For measurable \(E \subseteq \RR\) with \(m(E) < \infty\) and \(\eps > 0\), there is a finite union \(A\) of bounded open intervals with \(\|\chi_E - \chi_A\|_1 < 2\eps\).
Outer regularity gives an open \(U \supseteq E\) with \(m(U \setminus E) < \eps\), so \(m(U) < \infty\). \(U\) is a countable disjoint union of open intervals \(I_i\), each bounded because \(m(U) < \infty\), and \(\sum_i m(I_i) = m(U)\). Choose \(N\) with \(\sum_{i > N} m(I_i) < \eps\) and put \(A = \bigcup_{i \le N} I_i\). Then \(\|\chi_E - \chi_A\|_1 \le m(U \setminus E) + m(U \setminus A) < 2\eps\).
4.2. Q.E.D.
Given \(f \in L^1\) and \(\eps > 0\), step 3 gives a simple \(s = \sum_{i=1}^m a_i\chi_{E_i}\) with \(m(E_i) < \infty\) and \(\|f - s\|_1 < \eps\). Step 1 gives finite unions of intervals \(A_i\) with \(\|\chi_{E_i} - \chi_{A_i}\|_1 < \eps/(1 + \sum_i|a_i|)\), and \(\sum_i a_i\chi_{A_i}\) is a step function within \(2\eps\) of \(f\).
5. \(C_c^\infty(\RR^n)\) is dense in \(L^1(\RR^n)\).
5.1. For \(f \in L^1\), \(f_R = f\chi_{B(0,R)}\) satisfies \(\|f - f_R\|_1 \to 0\) as \(R \to \infty\).
\(|f - f_R| = |f|\chi_{\RR^n \setminus B(0,R)}\) tends to \(0\) pointwise and is bounded by \(|f| \in L^1\); apply dominated convergence.
5.2. Let \(\phi \in C_c^\infty(\RR^n)\) with \(\phi \ge 0\) and \(\int\phi = 1\), and \(\phi_\eps(x) = \eps^{-n}\phi(x/\eps)\). Then \(f_R \ast \phi_\eps \in C_c^\infty(\RR^n)\) and \(\|f_R \ast \phi_\eps - f_R\|_1 \to 0\) as \(\eps \to 0\).
Differentiation under the integral gives \(D^\alpha(f_R \ast \phi_\eps) = f_R \ast D^\alpha\phi_\eps\), which is continuous, and \(\supp(f_R \ast \phi_\eps) \subseteq \overline{B(0,R)} + \supp\phi_\eps\) is compact. The \(L^1\) convergence is the \(L^1\) convergence of approximations to the identity, which follows from continuity of translation in \(L^1\).
5.3. Q.E.D.
\(\|f - f_R \ast \phi_\eps\|_1 \le \|f - f_R\|_1 + \|f_R - f_R \ast \phi_\eps\|_1\), and steps 1 and 2 make both terms small.
6. Q.E.D.
Steps <1>1–<1>5 treat the five parts in order.