I am asked to show that the Lebesgue integral is invariant under translations. Specifically,
Let $(\mathbb{R}, \Sigma, \mu)$ be a measure space, and for any $f:\mathbb{R}\rightarrow\mathbb{R}$ define $f_a:\mathbb{R}\rightarrow\mathbb{R}$ by $f_a(x)=f(x-a)$, for all $x\in\mathbb{R}$. Assume that the measure $\mu$ is such that $\mu(A)=\mu(A+a)$, where $A+a=\{x+a:x\in A\}$, for $A\in\Sigma$ and $a\in\mathbb{R}$. Assume also that $A+a\in\Sigma$ for any $A\in\Sigma$ and $a\in\mathbb{R}$.
- Let $f:\mathbb{R}\rightarrow[0,\infty]$ be a simple, measurable function, $f=\sum_{j=1}^m f_j$, with $f_j=a_j\chi_{A_j}$, $a_j\ge0$ and $A_j\in\Sigma$ for all $1\le j \le m$. The sets $A_j$ are pairwise disjoint. Show that \begin{align} \int_{\mathbb{R}}f\, \mathrm{d}\mu = \int_{\mathbb{R}}f_a\,\mathrm{d}\mu.\end{align}
- Show that this also holds for any measurable $f:\mathbb{R}\rightarrow[0,\infty]$.
Now, I have attempted to solve the first part by writing $f_a$ as \begin{align} f_a(x) = f(x-a) &= \sum_{j=1}^m f_j(x-a) = \sum_{j=1}^m a_j\chi_{A_j}(x-a) = \sum_{j=1}^m a_j\chi_{A_j+a}(x), \end{align} and then computing \begin{align} \int_{\mathbb{R}}f(x)\,\mathrm{d}\mu(x) &= \int_{\mathbb{R}}\sum_{j=1}^m f_j(x)\,\mathrm{d}\mu(x)\\ &= \sum_{j=1}^m a_j\mu(A_j) = \sum_{j=1}^m a_j\mu(A_j+a) \\ &= \int_{\mathbb{R}}\sum_{j=1}^m f_{j,a}(x)\,\mathrm{d}\mu(x) = \int_{\mathbb{R}}f_a(x)\,\mathrm{d}\mu(x), \end{align} using the fact that $\mu(A_j)=\mu(A_j+a)$.
Now, does this make sense? Also, in order to generalize to arbitrary measurable, positive $f$, can I just use the fact that any measurable function is the pointwise limit of a sequence of simple functions, and apply the monotone convergence theorem?
Any help would be greatly appreciated!