Translation Invariance of Integral

• Mar 14th 2009, 12:39 PM
manjohn12
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 $\displaystyle |\mathcal{R}(g_k,P)-\mathcal{R}(f,P)| < \epsilon$?
• Mar 14th 2009, 06:08 PM
ThePerfectHacker
Quote:

Originally Posted by manjohn12
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.
• Mar 14th 2009, 06:29 PM
manjohn12
Quote:

Originally Posted by ThePerfectHacker
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 $\displaystyle I$?