6:13 AM
Let $I$ be an interval, with endpoints $a,b \in \Bbb{R}$. Define interval indicator function as follows:
$$1_I(x) = \begin{cases}c, x \in I \\\ 0, x \not\in I\end{cases}$$
Let $\mathcal{I} = \{I_j\}_{j \in J}$ be a pairwise disjoint collection of intervals and $J$ be a linearly ordered index set
such that $\bigcup \mathcal{I} = \Bbb{R}$
Now a step function $\Pi$ is defined as:
$$\Pi(x) = \bigcup_{I \in \mathcal{I}} 1_{I}$$
where $|\mathcal{I}| < \aleph_0$
We can now proceed to define a piecewise constant function as:
$$C(x) = \bigcup_{I \in \mathcal{I}} 1_{I}$$
where $|\mathcal{I}| \geq \aleph_0$
6
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 cl...
Using the above definition, a piecewise constant function is in general locally constant almost everywhere
(in the usual topology of the reals)
Therefore, the cantor function is also piecewise constant under this definition
as the points where it is not locally constant is a subset of the cantor set, which is measure zero, hence the cantor function is locally constant almost everywhere
In mathematics, the Smith–Volterra–Cantor set (SVC), fat Cantor set, or ε-Cantor set is an example of a set of points on the real line ℝ that is nowhere dense (in particular it contains no intervals), yet has positive measure. The Smith–Volterra–Cantor set is named after the mathematicians Henry Smith, Vito Volterra and Georg Cantor. The Smith-Volterra-Cantor set is topologically equivalent to the middle-thirds Cantor set.
== Construction ==
Similar to the construction of the Cantor set, the Smith–Volterra–Cantor set is constructed by removing certain intervals from the unit interval [0, 1].
The...
Therefore, a cantor function like function based on the fat cantor set is not piecewise constant since the points where it is not locally constant has positive measure, and hence violate the criteria to be almost
Proposition 1: Let $f$ be any real function that is continuous almost everywhere, and let $C$ be a piecewise constant function. Then $C(f)$ is piecewise constant in the subspace topology of the reals
Proof: Since $f$ is continuous almost everywhere, its range defines a collection of pairwise disjoint intervals such that their union is an interval $K$ minus a set of measure zero $Z$. Now the range of $C(f)$ is given by $C((K-Z) \cap \Bbb{R})$
Now, by definition, $C$ is locally constant almost everywhere. Let that set where it is not locally constant be $Z_2$. Thus the domain of $C$ consists of the reals partitioned into collection of intervals and $Z_2$
Meanwhile $K-Z$ is also a partition of an interval $K$ into collection of intervals and $Z$
The definition for step functions can be read here. I will refine that definition slightly and say that I am referring to functions that can have finite or infinite many pieces. This is mentioned in the first paragraph when comparing piecewise constant functions to step functions. I mostly just p...
Find an example of a continuous function $f: [0,1] \to [0,1]$ such that:
$f(0) = 0$
$f(1) = 1$
$f$ is locally constant almost everywhere
I know I need to come up with a function that fails to be locally constant on some subset of $[0,1]$ which has measure zero, but I'm struggling to see how a...
Inspired by L'Hosopital I started considering “ L'Hospital's equations “
Let $x$ be real and for all $ x > s $ we have $f(x),g(x)$ map to the reals.
Let $*^{(n)} $ denote the $n$ th derivative. $n$ is a strict positive integer.
Then L'Hospital's equations are
$$ \frac{f^{(n)}(x)}{g^{(n)}(x)...
