It is a classical fact that a $L^p(R^d)$ ($1\leq p<+\infty$) function can be approximated by step functions with compact support, but my question will be, can we require that the step function is monotonically increasing (suppose that $f>0$ a.e. or we can do it for $f^\pm$ respectively)?
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I googled and get no results. And my knowledge of proof of the classical fact can not modify easily to get a proof of my requirement. – van abel Jul 08 '14 at 09:02
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1What is your definition of a step function? Are we talking about simple functions (i.e. $\sum_{i=1}^n \alpha_i \chi_{A_i}$ with $A_i$ measurable) or about "Riemann" step functions, i.e. $\sum_{i=1}^n \alpha_i \chi_{A_i}$ with $A_i$ intervalls? – PhoemueX Jul 08 '14 at 09:26
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@PhoemueX Surely, he means the latter. If anybody ever says step function when they mean simple function, I would assume they misspoke. – Harald Hanche-Olsen Jul 08 '14 at 09:45
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I would not say that. For example Serge Lang uses the term "step map" to denote functions of the form $\sum_{i=1}^n \alpha_i \chi_{A_i}$ with $A_i$ measurable in his book "Real and Functional Analysis". – PhoemueX Jul 08 '14 at 09:48
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@PhoemueX The step function usually means simple function on Cubes( intervals in $R^1$ case). – van abel Jul 08 '14 at 13:22
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I expect you mean to ask if you can find a monotonically increasing sequence of step functions approximating the given function from below?
The answer is a resounding no. For a simple example, consider the characteristic function of a fat Cantor set.
Harald Hanche-Olsen
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Looking at non-negative unsigned Lebesgue integrable functions on $\mathbb{R}$,
- step function: taking finitely many values on finitely many intervals.
- simple function: taking finitely many values on finitely many Lebesgue measurable sets.
True: Both step functions and simple functions are dense in $L^1$. They can approximate any Lebesgue integrable function $f$ arbitrarily close using the integral norm $||\cdot||_{L^1}$.
True: For each unsigned Lebesgue integrable function $f$, there exists a sequence of monotone increasing simple functions $g_n$ such that $g_n\rightarrow f$ in $L^1$.
As Harald Olsen said, even step functions are dense in $L^1$, the statement about monotone is false. And here is the intuitive idea:
The pre-image of a step function has to be intervals. But given a Lebesgue measurable set $A$, if only using intervals, some times we can only approximate $A$ from the outside but not from the inside. That is $$\mu(A) = \inf\left\{ \sum_{n=1}^\infty |I_n| : A\subset \cup_n I_n \right\}$$ but $$\mu(A) \neq \sup\left\{ \sum_{n=1}^\infty |I_n| : A\supset \cup_n I_n \right\}.$$
Xiao
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I think you have some misunderstanding about the approximate, I think we should allow $g\leq f$ holds in the a.e. sense. In that case, we can take $f_k\equiv 1$ on $[0,1]$ as step function in your counterexample on $[0,1]$. – van abel Jul 08 '14 at 13:16
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@vanabel Yes, you are right. The fat Cantor set is the correct example. I was only thinking in term of measure while writing down this example. – Xiao Jul 08 '14 at 13:22
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By the way, Can you give some exactly reference or raise an answer here, of your second TRUE? I believe it will be useful to me, although the step function is replaced by simple function. – van abel Jul 08 '14 at 13:27
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@vanabel by the definition of unsigned integral $$\int f = \sup_{0\leq g\leq f \text{ a.e. }; \text{ simple}} \int g,$$ we know $f$ can be approximated from below by a sequence of simple functions $g_n$ in $L^1$ and pointwise a.e. Taking the maximum of the first $k$ simple functions will give you the monotone sequence of simple functions $h_k$. – Xiao Jul 08 '14 at 14:05
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