# Thread: Translation Invariance of Integral

1. ## Translation Invariance of Integral

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

Proof. Let $\epsilon >0$. Choose $\delta >0$ such that $\delta < \frac{\epsilon}{2}$. So basically we consider $\mathcal{R}(g_k, P) = \sum_{i=1}^{n} g_{k}(x_{i}^{*})(x_{i}-x_{i-1})$? And then show that $|\mathcal{R}(g_k,P)-\mathcal{R}(f,P)| < \epsilon$?

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

3. Originally Posted by ThePerfectHacker
Let $I = \smallint_a^b f$ we know that for any $\epsilon > 0$ there exists $\delta > 0$ so that $|\mathcal{R}(f,P) - I| < \epsilon$ where $P$ is any partition with $||P||<\delta$ and $\mathcal{R}(f,P)$ is any associated Riemann sum. Let $P_k$ be any partition of $[a-k,b-k]$ with $||P_k|| < \delta$. If $\sum_{j=1}^n g_k(t_j)(x_j-x_{j-1})$ is some Riemann sum then it is equal to $\sum_{j=1}^n f(x_j+k) [(x_j+k)-(x_{j-1}+k)]$, but this is a Riemann sum associated with $f$ where the partition norm is less than $\delta$. Thus, $\left| \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1}) - I \right| < \epsilon$, so we see that $g_k$ is integrable with the same value.
Would it still prove it if we said $\left| \sum_{j=1}^n g_k(t_j)(x_j-x_{j-1}) - \mathcal{R}(f,P) \right| < \epsilon$ instead of $I$?

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### translation invariance of lebesgue integral

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