Counter example that Lebesgue is translation invariant

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I seem to have a basic misunderstanding for what translation invariant means.

I thought it simply means that $\lambda(\varphi(x))=\lambda(\varphi(x+a))$ for some $a\in \mathbb{R}^n$. Here $\varphi$ is in $C_c(\mathbb{R}^n)$


But, this example doesn't work:

$\int_{[0,1]} x^2 d\lambda = \int_0^1 x^2 dx = \frac{1}{3} \neq \int_0^1 (x+1)^2 dx = \int_{[0,1]} (x+1)^2 d\lambda$

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Okay I got it:

$\varphi(x) = \{x^2$ if $x\in [0,1], \,\,0 $ else

Then: $\int \varphi(x) dx = \frac{1}{3} = \int \varphi(x+1) dx = \int_{-1}^{0} (x+1)^2 dx = \frac{1}{3}$