Let $\lambda=l^*$ denote Lebesgue measure on $\Bbb R$, and let $A$ be a Lebesgue measurable set with $\lambda(A)\lt +\infty$. Show that if $\epsilon \gt0$, there exists an open set which is the union of a finite number of open intervals such that $$||\chi_A-\chi_G||_1=|\lambda(A)-\lambda(G)|\lt \epsilon$$

Also, if $\epsilon \gt0$ there exists a continuous function $f$ such that $$||\chi_A-f||_1=\int |\chi_A-f| d\lambda\lt \epsilon$$

I already showed that $$l^*(A)=inf\{l^*(G):A\subseteq G, \text{G is open}\}$$ $$l^*(A)=sup\{l^*(K):K\subseteq A, \text{K is compact}\}$$, but I'm not sure if these could help prove the above proposition. Could someone provide a complete proof please? Thanks.

  • 1
    $\begingroup$ No complete proof, so just a comment. Find $K\subset A\subset G$ approximating $A$. $G$ is a countable union of intervals, and a union $G'$ of finitely many suffice to cover $K$. It should work to use $G'$ to construct your function, say, piecewise linearly. $\endgroup$ – neth Apr 20 '16 at 17:59
  • $\begingroup$ @neth Thanks, I tried to work on the ideas but still having difficulties coming up with such $G$ and $K$, could you provide further details or some explicit steps please? $\endgroup$ – EmmaJ Apr 21 '16 at 6:33

Here is a sketch. Fix $\epsilon>0$. Because you know $$\lambda(A)=\inf\{\lambda(G):A\subset G\mbox{ open}\}$$ and $$\lambda(A)=\sup\{\lambda(K):A\supset K\mbox{ compact}\},$$ you know there is $K\subset A\subset G$ with $K$ compact and $G$ open such that $\lambda(A\setminus K)<\frac{\epsilon}{2}$ and $\lambda(G\setminus A)<\frac{\epsilon}{2}$.

Because $G\subset\mathbb{R}$ is open, we can write $G$ as a disjoint union of open intervals, $G=\bigcup_{i\in I}U_i$. So $\{U_i\}_{i\in I}$ is an open cover of $K$, whence by compactness of $K$ it has a finite subcover $\{U_1,\ldots,U_n\}$. Set $H=U_1\cup\cdots\cup U_n$. Then as $K\subset A,H\subset G$, you know $\lambda(K)\leq \lambda(A),\lambda(H)\leq G$, so that $|\lambda(A)-\lambda(H)| < \epsilon$. (In fact, $\lambda(A\Delta H)<\epsilon$, where $\Delta$ is symmetric difference.)

To define $f$ in the case $A=(a,b)$, simply let $f_\delta$ be zero in $(-\infty,a-\delta]$, linearly rise from $0$ to $1$ in $[a-\delta,a+\delta]$, be $1$ on $[a+\delta,b-\delta]$, go down to $0$ in $[b-\delta,b+\delta]$, and stay $0$ forevermore. As you take $\delta$ as small as you'd like, $\|\chi_A-f_\delta\|_1$ gets as small as you'd like. For general $A$, approximate $A$ with a finite union of disjoint intervals, and approximate it by a finite sum of such $f$'s with an $\frac{\epsilon}{n}$ argument.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.