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Chapter 11: Problem 2

Find a step function \(s(x)\) that approximates \(f(x)=x^{2}\) uniformly to within \(\varepsilon>0\) on \([0,1]\), in the sense that \(|f(x)-s(x)|<\varepsilon\) cierywhere in \([0,1]\).

Short Answer

Expert verified
A step function which approximates \(f(x)=x^2\) uniformly to within \(\varepsilon>0\) on \([0,1]\) can be constructed by partitioning the interval into sufficiently small subintervals of width \(\Delta x = \sqrt\varepsilon\) and then defining the step function to be \(s(x) = \bar{x}^2\) on each subinterval, where \(\bar{x}\) is any number in that subinterval. The deviation between \(f(x)\) and \(s(x)\) will be less than \(\varepsilon\).

Step by step solution

01

Define the function

Start by defining the given function \(f(x)=x^2\) and the boundary interval \([0,1]\).
02

Partition the interval

Next, partition the interval \([0,1]\) into subintervals of width \(\Delta x\). The value of \(\Delta x\) is determined by the maximum deviation allowed, \(\varepsilon\). It is crucial to ensure that \(\Delta x\) is small enough so that the maximum difference between \(f(x)\) and any constant \(\bar{x}^2\) within each subinterval \([x-\Delta x, x]\) doesn't exceed \(\varepsilon\). It will depend on the specific number \(\varepsilon\) but a reasonable choice could be \(\Delta x = \sqrt{\varepsilon}\).
03

Define the step function

Now, define the step function \(s(x)\). Set \(s(x) = \bar{x}^2\) on each subinterval, where \(\bar{x}\) is any number in the subinterval. It can be chosen as the midpoint of each subinterval or the left or right endpoint - as long as it is inside the subinterval. Within each subinterval, the step function resolves to a single constant value. Since \(\Delta x\) is sufficiently small, the value of \(s(x)\) should not deviate from \(f(x)\) by more than \(\varepsilon\) in each subinterval.
04

Uniform approximation

Ensure that \(s(x)\) uniformly approximates \(f(x)=x^2\), in the sense that \(|f(x)-s(x)|<\varepsilon\) for every \(x\) in \([0,1]\). By construction, the maximum difference occurs at the boundaries of the subintervals and doesn't exceed \(\varepsilon\), thus \(s(x)\) is a successful approximation.

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Most popular questions from this chapter

A measure is said to be complete if every subset of a sct of measure zero is measurable. Show that if \(A \subset \mathbb{R}\) is a set of outer measure zero, \(\mu^{*}(A)=0\), then \(A\) is Lebesgue measurable and has measur zero. Hence shew that Lebesgue measure is complcte.

If \(E\) is bounded and there exists an interval \(I \supset E\) such that $$ \mu^{*}(I)=\mu^{*}(I \cap E)+\mu^{*}(I-E) $$ then this holds for all intervals, possibly even those overlapping \(E\).

If \(\mu^{*}(N)=0\) show that for any set \(E, \mu^{*}(E \cup N)=\mu^{*}(E-N)=\mu^{*}(E)\). Hence show that \(E \cup N\) and \(E-N\) are Lebesgue measurable if and only if \(E\) is measurable.

Let \(f: X \rightarrow \mathbb{R}\) and \(g: X \rightarrow \mathbb{R}\) be mcasurable functicns and \(E \subset X\) a measurable set. Show that $$ h(x)= \begin{cases}f(x) & \text { if } x \in E \\ g(x) & \text { if } x \not E\end{cases} $$ is a measurable function on \(X\).

The inner measure \(\mu_{*}(E)\) of a set \(E\) is defined as the least upper bound of the measures of all measurable subsets of \(E\). Show that \(\mu_{*}(E) \leq \mu^{*}(E)\). For any open set \(U \supset E\), show that $$ \mu(U)=\mu_{*}(U \cap E)+\mu^{*}(U-E) $$ and that \(E\) is measurable with finite measure if and only if \(\mu_{*}(E)=\mu^{*}(E)<\infty\).
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