Lebesgue integral of f(x) over R is equal to Lebesgue integral of f(x+t) over R

Question

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.

Answer 1

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:

1. 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.

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

3. 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.

Answer 2

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*}