Lebesgue integral of f(x) over R is equal to Lebesgue integral of f(x+t) over R What is the most natural way to show that for Lebesgue-integrable function $f$ we have
\begin{equation}
\int f(x)dx=\int f(x+t)dx?
\end{equation}
Is the following approach correct?
Let $\epsilon>0$. We can find simple function $\psi$ such that $\int|f(x)-\psi(x)|<\epsilon/2$. Then $\psi(x+t)$ is simple function such that $\int|f(x+t)-\psi(x+t)|dx<\epsilon/2$ and $\int\psi(x)dx=\int\psi(x+t)dx$. From here we have
\begin{equation}
\left|\int f(x+t)dx-\int f(x)dx\right|\leq\int|f(x)-\psi(x)|dx+\left|\int\psi(x)dx-\int\psi(x+t)dx\right|+\int|f(x+t)-\psi(x+t)|dx<\epsilon,
\end{equation}
end we are done.
 A: So far you have proved that have
\begin{equation}
\int f(x)dx \to \int f(x+t)dx
\end{equation}
as $ t \to 0.$ But this does not prove what you're trying to prove. If are trying to prove that the Lebesgue integral is translation invariant, do the following:


*

*Show that the equality is true for any indicator function, which is obviously true because of the definition of Lebesgue integral and translation invariance of Lebesgue measure.

*Using finite linearity of Lebesgue integral, show that the result holds for any simple function.

*Without loss of generality assume that $f$ is non-negative, otherwise, write $f= f^{+}-f^{-}.$  Let's say $f$ is non-negative. Then there exists a monotone increasing sequence of simple function ${s_{n}}$ such that $s_{n} \to f.$ Then apply MCT.
Then you're done. Let me know if you have any questions.
Edit: Since your $\epsilon$ is arbitrary and $t$ is independent of $\epsilon,$ your proof should be good. Sorry for my irresponsible comments.
A: The main idea lying under the equality you're interested in, is that the set $x\mapsto f(x)$ and $x\mapsto f(x+t)$ run over, is the same: the whole $\Bbb R$.
This is because $x\mapsto \phi(x):=x+t$ is a diffeomorphism of $\Bbb R$, for each fixed $t$.
Moreover, the way these two functions runs over $\Bbb R$ is the same: this is expressed by the fact the derivative of $x\mapsto \phi(x)$ is identically 1.
What I wrote is resumed by the change of variable theorem for integrals:
\begin{align*}
\int_{\Bbb R}f(x+t)dx
&=\int_{\Bbb R}f(\phi(x))dx\\
&=\int_{\underbrace{\phi(\Bbb R)}_{=\Bbb R}}f(y)\cdot\underbrace{(\phi^{-1})'(y)}_{\equiv1} dy\\
\end{align*}
