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Title: Definition of Absolute Continuity
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Series: Absolutely Continuous Functions
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YouTube-Title: Absolutely Continuous Functions 1 | Definition of Absolute Continuity
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Forum: Ask a question in Mattermost
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Quiz: Test your knowledge
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Subtitle on GitHub: ac01_sub_eng.srt missing
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Download bright video: Link on Vimeo
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Definitions in the video: absolutely continuous functions, uniformly continuous functions
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. How to define continuity at the point $x$?
A1: $ \forall \varepsilon > 0 ~ \exists \delta > 0 ~\forall \tilde{x} \in \mathbb{R} : $ $ ~ | x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| < \varepsilon$
A2: $ \forall \varepsilon > 0 ~\exists \delta > 0 ~\forall \tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| > \varepsilon$
A3: $ \forall \varepsilon > 0 ~\exists \delta > 0 ~\forall \tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | > \delta \Rightarrow |f(x) - f(\tilde{x})| > \varepsilon$
A4: $ \forall \varepsilon > 0 ~\forall \delta > 0 ~\exists \tilde{x} \in \mathbb{R}: $ $ ~| x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| > \varepsilon$
A5: $ \forall \varepsilon > 0 ~\forall \delta > 0 ~\exists \tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| < \varepsilon$
A6: $ \forall \varepsilon > 0 ~\forall \delta > 0 ~\exists \tilde{x} \in \mathbb{R}: $ $ ~| x - \tilde{x} | > \delta \Rightarrow |f(x) - f(\tilde{x})| < \varepsilon$
Q2: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. How to define uniform continuity?
A1: $ \forall \varepsilon > 0 ~ \exists \delta > 0 ~\forall x,\tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| < \varepsilon$
A2: $ \forall \varepsilon > 0 ~\exists \delta > 0 ~\forall x,\tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| > \varepsilon$
A3: $ \forall \varepsilon > 0 ~\exists \delta > 0 ~\forall x,\tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | > \delta \Rightarrow |f(x) - f(\tilde{x})| > \varepsilon$
A4: $ \forall \varepsilon > 0 ~\forall \delta > 0 ~\exists x,\tilde{x} \in \mathbb{R}: $ $ ~| x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| > \varepsilon$
A5: $ \forall \varepsilon > 0 ~\forall \delta > 0 ~\exists x,\tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | < \delta \Rightarrow |f(x) - f(\tilde{x})| < \varepsilon$
A6: $ \forall \varepsilon > 0 ~\forall \delta > 0 ~\exists x,\tilde{x} \in \mathbb{R}: $ $ ~ | x - \tilde{x} | > \delta \Rightarrow |f(x) - f(\tilde{x})| < \varepsilon$
Q3: Let $f: [a,b] \rightarrow \mathbb{R}$ be an absolutely continuous function. What is not possible?
A1: We find an $\varepsilon > 0$ such that for all $\delta > 0$ there are pairwise disjoint intervals $(x_k, \tilde{x}_k)$ with the property:
$$ \sum_{k=1}^m | x_k - \tilde{x}_k | < \delta \text{ and }\sum_{k=1}^m | f(x_k) - f(\tilde{x}_k) | \geq \varepsilon \ $$A2: We find an $\varepsilon > 0$ such that for all $\delta > 0$ there are pairwise disjoint intervals $(x_k, \tilde{x}_k)$ with the property:
$$ \sum_{k=1}^m | x_k - \tilde{x}_k | < \delta \Rightarrow \sum_{k=1}^m | f(x_k) - f(\tilde{x}_k) | < \varepsilon \ $$A3: For all $\varepsilon > 0$ there is a $\delta > 0$ such that for all pairwise disjoint intervals $(x_k, \tilde{x}_k)$ we have the property:
$$ \sum_{k=1}^m | x_k - \tilde{x}_k | < \delta \Rightarrow \sum_{k=1}^m | f(x_k) - f(\tilde{x}_k) | < \varepsilon \ $$A4: For all $\varepsilon > 0$ there is a $\delta > 0$ such that for all pairwise disjoint intervals $(x_k, \tilde{x}_k)$ we have the property:
$$ \sum_{k=1}^m | x_k - \tilde{x}_k | < \delta \text{ or } \sum_{k=1}^m | f(x_k) - f(\tilde{x}_k) | < \varepsilon \ $$ -
Date of video: 2026-02-16
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Last update: 2026-02
