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Thread: Translation Invariance of Integral

  1. #1
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    Translation Invariance of Integral

    Theorem. Let $\displaystyle f $ be a real-valued function that is Riemann integrable on $\displaystyle [a,b] $. For a fixed $\displaystyle k \in \mathbb{R} $, let $\displaystyle g_k $ be the function $\displaystyle g_{k}(x) = f(x+k) $ where the domain of $\displaystyle g $ is appropriately chosen so that $\displaystyle x $ is in the domain of $\displaystyle g_k $ if and only if $\displaystyle x+k $ is in the domain of $\displaystyle f $. Then $\displaystyle g_k $ is Riemann integrable on $\displaystyle [a-k, b-k] $ and $\displaystyle \smallint_{a-k}^{b-k} g_k = \smallint_{a}^{b} f $.


    Proof. Let $\displaystyle \epsilon >0 $. Choose $\displaystyle \delta >0 $ such that $\displaystyle \delta < \frac{\epsilon}{2} $. So basically we consider $\displaystyle \mathcal{R}(g_k, P) = \sum_{i=1}^{n} g_{k}(x_{i}^{*})(x_{i}-x_{i-1}) $? And then show that $\displaystyle |\mathcal{R}(g_k,P)-\mathcal{R}(f,P)| < \epsilon $?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Theorem. Let $\displaystyle f $ be a real-valued function that is Riemann integrable on $\displaystyle [a,b] $. For a fixed $\displaystyle k \in \mathbb{R} $, let $\displaystyle g_k $ be the function $\displaystyle g_{k}(x) = f(x+k) $ where the domain of $\displaystyle g $ is appropriately chosen so that $\displaystyle x $ is in the domain of $\displaystyle g_k $ if and only if $\displaystyle x+k $ is in the domain of $\displaystyle f $. Then $\displaystyle g_k $ is Riemann integrable on $\displaystyle [a-k, b-k] $ and $\displaystyle \smallint_{a-k}^{b-k} g_k = \smallint_{a}^{b} f $.
    Let $\displaystyle I = \smallint_a^b f$ we know that for any $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that $\displaystyle |\mathcal{R}(f,P) - I| < \epsilon$ where $\displaystyle P$ is any partition with $\displaystyle ||P||<\delta$ and $\displaystyle \mathcal{R}(f,P)$ is any associated Riemann sum. Let $\displaystyle P_k$ be any partition of $\displaystyle [a-k,b-k]$ with $\displaystyle ||P_k|| < \delta$. If $\displaystyle \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1})$ is some Riemann sum then it is equal to $\displaystyle \sum_{j=1}^n f(x_j+k) [(x_j+k)-(x_{j-1}+k)]$, but this is a Riemann sum associated with $\displaystyle f$ where the partition norm is less than $\displaystyle \delta$. Thus, $\displaystyle \left| \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1}) - I \right| < \epsilon$, so we see that $\displaystyle g_k$ is integrable with the same value.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle I = \smallint_a^b f$ we know that for any $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that $\displaystyle |\mathcal{R}(f,P) - I| < \epsilon$ where $\displaystyle P$ is any partition with $\displaystyle ||P||<\delta$ and $\displaystyle \mathcal{R}(f,P)$ is any associated Riemann sum. Let $\displaystyle P_k$ be any partition of $\displaystyle [a-k,b-k]$ with $\displaystyle ||P_k|| < \delta$. If $\displaystyle \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1})$ is some Riemann sum then it is equal to $\displaystyle \sum_{j=1}^n f(x_j+k) [(x_j+k)-(x_{j-1}+k)]$, but this is a Riemann sum associated with $\displaystyle f$ where the partition norm is less than $\displaystyle \delta$. Thus, $\displaystyle \left| \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1}) - I \right| < \epsilon$, so we see that $\displaystyle g_k$ is integrable with the same value.
    Would it still prove it if we said $\displaystyle \left| \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1}) - \mathcal{R}(f,P) \right| < \epsilon$ instead of $\displaystyle I $?
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