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Let $ \ f \ $ be integrable on $ \ E \ $.

Prove $ \ \int_E f(x)dx=\int_E f(x+t)dx \ $ where $ \ E=(-\infty, \infty) \ $ and $ \ E \ $ is measurable

Answer:

Since $ \ f \ $ is integrable , we have

$ \int_E f(x)dx <\infty \ $

Let $ E_1,E_2,......,E_N \ $ are the disjoint partition (interval) of $ E \ $ such that

$ f(x)=a_i , \ \ if \ \ x\in E_i \ $

Let $ \ \mu \ $ be the measure on E , then $ \ \mu(E)=|E| \ $

Then,

$ \int_E f(x)dx=\sum_{i=0}^{N} a_i |E_i|=\int_E f(t+x)dx \ \\ \\ \Rightarrow \int_E f(x)dx= \int_E f(t+x)dx $

(Proved)

But this concept works if $ \ f \ $ be a simple function.

I need help to overcome this.

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Assume first that $f\geq 0$, then find an increasing sequence $0\leq\varphi_{n}\leq f$ of simple functions such that $\varphi_{n}\uparrow f$ a.e., and we have by Monotone Convergence Theorem that \begin{align*} \int f=\lim_{n}\int\varphi_{n}=\lim_{n}\int\varphi_{n}(\cdot+t)=\int f(\cdot+t). \end{align*} For general functions $f$, split $f=f^{+}-f^{-}$ and tackle separately to $f^{+}$ and $f^{-}$.

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  • $\begingroup$ please do a little more. I have still problem to understand $\endgroup$ – M. A. SARKAR Apr 10 '18 at 19:27
  • $\begingroup$ Do you mean $ \int _E f(x)dx=\lim_{n} \int_E \phi_n(x)dx=\lim_{n} \int_E \phi(x+t)dx=\int_E f(x+t)dx \ $ ? $\endgroup$ – M. A. SARKAR Apr 10 '18 at 19:30
  • $\begingroup$ So we have to show this for $ \ f^{+} \ $ and $ \ f^{-} \ $ separately. Then we have to combine the result. Is it > $\endgroup$ – M. A. SARKAR Apr 10 '18 at 19:37
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    $\begingroup$ Yes, you are right again. $\endgroup$ – user284331 Apr 10 '18 at 19:44

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