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Definitions: An integrable simple function $f = a_1 \chi_{A_1} + \ldots + a_n \chi_{A_n},$ where $\chi_{A_k}$ is an indicator function and the $A_j$'s are disjoint measurable sets of finite measure. A step function is a piecewise constant function with a finite number of pieces (basically it is an ISF with intervals as the sets).

Prove that for any ISF $f$ and $\epsilon > 0,$ there is a step function $g$ such that $\displaystyle\int |f-g| < \epsilon.$

The question says to use the approximation property for measurable sets (measurable sets can be approximated by open and closed sets) and the fact that $m(A\triangle B) = \displaystyle\int |\chi_A - \chi_B|,$ where $A\triangle B$ is the symmetric difference.

Basically, I'm having trouble finding a finite set of intervals for this approximation - any advice would be appreciated.

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up vote 5 down vote accepted

I am assuming that your domain is $\mathbb{R}$ with the Lebesgue measure.

As is typical with these sort of things, first we try to approximate a single indicator function $1_A$.

Let $\epsilon>0$. If $A$ has finite measure, then there exists an open $U$ such that $m (U \setminus A) < \frac{1}{2} \epsilon$.

Any open $U \subset \mathbb{R}$ can be written as $U = \cup_{k=1}^\infty I_k$, where $\{I_k\}$ is a countable collection of disjoint open intervals. Since the $I_k$ are disjoint we have $m U = \sum_{k=1}^\infty m I_k$. Choose a finite collection $\{I_k\}_{k \in J}$ such that $mU -\sum_{k \in J} m I_k = m ( U\setminus \cup_{k \in J} I_k) < \frac{1}{2} \epsilon$.

Since $\cup_{k \in J} I_k \subset U$ and $A \subset U$, it follows that $\cup_{k \in J} I_k \setminus A \subset U \setminus A$, and $A \setminus \cup_{k \in J} I_k \subset U \setminus \cup_{k \in J} I_k$, hence $A \triangle \cup_{k \in J} I_k \subset (U \setminus A )\cup (U \setminus \cup_{k \in J} I_k)$, from which it follows that $m(A \triangle \cup_{k \in J} I_k ) < \epsilon$. Hence $\int |1_A -1_{\cup_{k \in J} I_k}| = m(A \triangle \cup_{k \in J} I_k ) < \epsilon$. Hence the characteristic function of any $A$ with finite measure can be approximated by the characteristic function of a finite collection of disjoint intervals.

I need some notation: Let $B$ ($=\cup_{k \in J} I_k$ above) be the finite union of intervals that approximates $A$ as above.

Suppose we are given $f=\sum_k a_k 1_{A_k}$, and choose $\epsilon>0$. Now choose the unions of intervals $B_k$ such that $m(A_k \triangle B_k) < \frac{1}{\sum_k |a_k|} \epsilon$. Then $$\int |f-\sum_k a_k 1_{B_k}| = \int |\sum_k a_k (1_{A_k}-1_{B_k})| \leq \sum_k |a_k| \int|1_{A_k}-1_{B_k}| < \epsilon$$ Since $\sum_k a_k 1_{B_k}$ is a step function, we are finished.

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thanks - I don't know how I missed this argument... must be lack of sleep – cats Feb 15 '13 at 8:17
@copper.hat I think that you are wrong here: $ mU - \sum\limits_{k \in J} {mI_k } = m\left( {U\backslash \bigcap\limits_{k \in J} {I_k } } \right) $ The real equality is this : $ mU - \sum\limits_{k \in J} {mI_k } = m\left( {U\backslash \bigcap\limits_{k \in J} {I_k ^c } } \right) $ – Miguel Jun 3 '13 at 19:44
@Miguel: Thanks for bringing this to my attention. I should have had $\cup_{k \in J} I_k$ instead of $\cap_{k \in J} I_k$ everywhere. – copper.hat Jun 3 '13 at 21:14
@copper.hat Thanks to you too copper.hat your idea to prove the aproximation was very useful for me :D! – Miguel Jun 3 '13 at 21:20
Delighted to be able to help. (I shouldn't take credit though, the argument above is fairly standard in measure theory.) – copper.hat Jun 3 '13 at 21:32

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