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Let $L$ be the space of piecewise-constant functions on $[0,1]\subset \mathbb{R}$ equipped with the supremum norm (i.e. step functions).

What is the completion of this space? We discussed in my class that all metric spaces have (unique) completion, but the proof of existence using equivalence classes doesn't give much mechanism to actually compute the completion.

Intuitively, the space should contain all continuous functions, but it should be strictly larger as we can have a sequence $f_n \in L$ which is just the constant sequence of a function which is discontinuous.

My guess would be $C[0,1] \cup \{f | f $ is piecewise constant$ \}$, and I could probably show that $L$ is dense in this space, but I'm not sure how I'd go about showing that it is complete.

Notably these functions are all integrable, so perhaps that's the completion - but we haven't discussed integrals yet in our course which makes me a bit hesitant.

Ali
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  • What exactly do you mean by piecewise constant functions? Step functions or simple functions? – mechanodroid Sep 07 '17 at 23:42
  • Step functions. –  Sep 07 '17 at 23:45
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    Remember the following fact from linear algebra: the union of two subspaces of a vector space will be a subspace if and only if one is a subset of the other. So, given the completion will also be a vector space, and will be strictly larger than both $C[0, 1]$ and the set of piecewise constant functions, your guess can't possibly be the answer. – Theo Bendit Sep 08 '17 at 00:13
  • I've just figured out that the Riemann integrable functions are not the completion. Take, for example, the indicator function for the range of the sequence $1/n$. This function is Riemann integrable, but cannot be approximated by piecewise constant functions. – Theo Bendit Sep 08 '17 at 00:41
  • The space of Riemann integrable functions seems to be complete, though. – mechanodroid Sep 08 '17 at 07:34
  • @mechanodroid no it isnt. You can find a Cauchy sequence converging to the Dirichlet function. – Vim Sep 21 '17 at 02:02
  • @Vim I found several references to this result: here and here. How would your sequence look like? If $\mathbb{Q} = (q_n){n=1}^\infty$, then $$\chi{{q_1, \ldots, q_n}} \xrightarrow{n\to\infty} \chi_{\mathbb{Q}}$$ only pointwise, not uniformly. – mechanodroid Sep 21 '17 at 12:14
  • @mechanodroid alrite, it's my bad. I was thinking about Riemann intégrable functions under the $L^1$ metric. But in our problem it is the uniform metric of course. – Vim Sep 21 '17 at 12:17
  • @Vim Yeah, the completion of step functions under $|\cdot|1$ is $L^1[0,1]$. For $|\cdot|\infty$, the step functions are not dense in $R[0,1]$ even though the space is complete. – mechanodroid Sep 21 '17 at 12:29

1 Answers1

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The completion is the space of regulated functions.

Dap
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