# Thread: Analysis - c-translate of an integrable function is integrable

1. ## Analysis: c-translate of an integrable function is integrable

I have difficulty proving this simple result: If $\displaystyle f\in \mathcal{R}[a,b]$, then $\displaystyle g(y):=f(y-c)\in \mathcal{R}[a+c,b+c]$ for $\displaystyle c\in \Bbb R$.
I want to use Cauchy condition to prove this. (If you know other simple proofs, please tell me.)

My attempt: Let $\displaystyle \varepsilon>0$. Since $\displaystyle f\in \mathcal{R}[a,b]$, there exists a partition of $\displaystyle [a,b]$, $\displaystyle P=\{a=x_0,x_1,\cdots,x_n=b\}$ such that $\displaystyle |U(f,P)-L(f,P)|<\varepsilon$. Pick $\displaystyle P_c=\{a+c=x_0+c,x_1+c,\cdots,x_n+c=b+c\}$. Then $\displaystyle |U(g,P_c)-L(g,P_c)|=|U(f,P_c)-L(f,P_c)|$. I have no idea what to do now. How can I compare this with $\displaystyle |U(f,P)-L(f,P)|$? Those are completely different partition points.

Can someone also explain the arguments in this alternative proof?