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Show that \(f\) is absolutely continuous (in the weaker sense) on \([a, b]\) if for every \(\varepsilon>0\) there is \(\delta>0\) such that $$ \begin{array}{c} \sum_{i=1}^{m}\left|f\left(t_{i}\right)-f\left(s_{i}\right)\right|<\varepsilon \text { whenever } \sum_{i=1}^{m}\left(t_{i}-s_{i}\right)<\delta \text { and } \\\ a \leq s_{1} \leq t_{1} \leq s_{2} \leq t_{2} \leq \cdots \leq s_{m} \leq t_{m} \leq b \end{array} $$ (This is absolute continuity in the stronger sense.)

Short Answer

Expert verified
\(f\) is absolutely continuous if small intervals produce small total changes.

Step by step solution

01

Understand the Hypothesis

We start with the given hypothesis that for an arbitrary small positive number \( \varepsilon \), there exists a small positive number \( \delta \) such that for any sequence of intervals \([s_i, t_i]\) within \([a, b]\), the inequality \( \sum_{i=1}^{m} |f(t_i) - f(s_i)| < \varepsilon \) holds whenever \( \sum_{i=1}^{m} (t_i - s_i) < \delta \). This suggests \( f \) appreciates in a controlled manner parallel to the length of the interval sums.
02

Define Absolute Continuity

A function \( f \) is absolutely continuous on \([a, b]\) if for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) so that whenever a finite collection of non-overlapping intervals \([x_i, y_i]\) satisfy \( \sum_{i=1}^{n} (y_i - x_i) < \delta \), it follows that \( \sum_{i=1}^{n} |f(y_i) - f(x_i)| < \varepsilon \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Epsilon-Delta Definition
The Epsilon-Delta definition is a key concept used to describe absolute continuity in functions. Essentially, it sets the rules for how changes in the interval size correlate to changes in the function value.

When we say that a function \( f \) is absolutely continuous, we mean that for every small number \( \varepsilon > 0 \), there exists another small number \( \delta > 0 \) such that for any chosen sequence of intervals with total length less than \( \delta \), the total variation in the function's values is less than \( \varepsilon \).

This precise relationship ensures that the function does not wildly oscillate within any given segment of its domain. It's a precise way of boxing the function's behavior into predictable patterns as intervals narrow down.
  • This establishments helps us ensure function behavior closely resembles that of a straight line over very small segments of its domain.
  • It helps in understanding whether a function maintains a controlled manner of increase or decrease regardless of the segment considered.
Non-overlapping Intervals
Non-overlapping intervals form a crucial part of the discussion around absolute continuity. Imagine having a series of ropes laid out without touching or crossing each other; these represent non-overlapping intervals.

In mathematical terms, these intervals ensure that each pair \([x_i, y_i]\) does not share any common points with another pair. This is significant in dealing with absolute continuity since it tests whether a function can limit its total variation across separate and independent pockets of its domain.
  • This helps in preventing double counting of variations when evaluating over multiple sections.
  • For a function to be absolutely continuous, the summed variations in these non-overlapping intervals must be less than \( \varepsilon \) when their length sum is less than \( \delta \).
This idea helps mathematicians and students confirm that the function’s increase or decrease is under control at every possible point within the domain, without interference from nearby checks.
Controlled Variation
The concept of controlled variation is intertwined with how a function's values change across its domain. In simple terms, if a function has controlled variation, its values will change in a manner that is directly proportional to the change in the interval considered.

In context to our problem with absolute continuity, the controlled variation implies that the function \( f \) showcases limited and predictable changes over a given group of subintervals. This means for small changes in interval sizes, the changes in function values are appropriately dampened and controlled to not exceed any small chosen \( \varepsilon \).
  • It prevents the function from making large jumps as you move along its graph.
  • Ensuring controlled variation across any part of the segment \([a, b]\) signifies a stable behavior pattern of the function.
This concept provides confidence in the otherwise unpredictable motions of some mathematical functions, assigning a precise boundary to their behavior in specific domains.

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Most popular questions from this chapter

Prove that if \(\int f\) exists on each \(I_{n}=\left[a_{n}, b_{n}\right],\) where $$ a_{n+1} \leq a_{n} \leq b_{n} \leq b_{n+1}, \quad n=1,2, \ldots $$ then \(\int f\) exists on $$ I=\bigcup_{n=1}^{\infty}\left[a_{n}, b_{n}\right] $$ itself an interval with endpoints \(a=\inf a_{n}\) and \(b=\sup b_{n}, a, b \in E^{*}\). [Hint: Fix some \(c \in I_{1}\). Define $$ H_{n}(t)=\int_{c}^{t} f \text { on } I_{n}, n=1,2, \ldots $$ Prove that $$ (\forall n \leq m) \quad H_{n}=H_{m} \text { on } I_{n}\left(\text { since }\left\\{I_{n}\right\\} \uparrow\right) . $$ Thus \(H_{n}(t)\) is the same for all \(n\) such that \(t \in I_{n},\) so we may simply write \(H\) for \(H_{n}\) on \(I=\bigcup_{n=1}^{\infty} I_{n} .\) Show that \(H=\int f\) on all of \(I ;\) verify that \(I\) is, indeed, an interval.]

In the following cases show that \(V_{f}[I]=+\infty,\) though \(f\) is bounded on I. (In case (iii), \(f\) is continuous, and in case (iv), it is even differentiable on \(I .)\) $$ \text { (i) For } I=[a, b](a

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