1.7 KiB
Math4121 Lecture 37
Extended fundamental theorem of calculus with Lebesgue integration
Density of continuous functions
Lemma:
Let K\subseteq U be bounded sets in \mathbb{R}, K is closed and U is open. Then there is a continuous function g such that \chi_K\leq g\leq \chi_U.
Proof in homework.
Hint: Consider the basic intervals cases.
Theorem for continuous functions
Let f be integrable. For each \epsilon>0, there is a continuous function g:\mathbb{R}\to\mathbb{R} such that \int_{\mathbb{R}}|f-g|dm<\epsilon.
Proof:
First where f=\chi_S for some bounded means set S. then extended to all f integrable.
First, assume f=\chi_S. Let \epsilon>c, we can find K\subseteq S\subseteq U. and K is closed and U is open such that (by definition of Lebesgue outer measure)
m(K)+\frac{\epsilon}{2}>m(S)>m(U)-\frac{\epsilon}{2}
In particular, m(U\setminus K)=m(U)-m(K)<\epsilon.
By lemma, there is a continuous function g such that \chi_K\leq g\leq \chi_U.
So
\int_E |\chi_S -g|dm=\int_{U\setminus K} |\chi_S -g|dm\leq m(U\setminus K)<\epsilon
For the general case,
By the Monotone Convergence Theorem (use |f|\chi_{[-N,N]} to approximate |f|), we can find N large such that
\int_{E_N^c}|f|dm<\frac{\epsilon}{2}
where E_N=E\cap [-N,N].
Notice that by the definition of Lebesgue integral, \int f^+ dm=\sup\{\int \phi^+ dm:\phi\text{ is simple and } \phi\leq f^+\} and \int f^- dm=\sup\{\int \phi^- dm:\phi\text{ is simple and } \phi\leq f^-\}.
By considering f^+ and f^- separately, we can find a simple function \phi such that
\int_{E_N} |f-\phi|dm<\frac{\epsilon}{3}
QED