You are just trying to show that it is integrable.
We know $g$ is measurable because the functions $x \mapsto x^{-{3 \over 2} } 1_{(0,1]}(x)$ and $f$ are.
Since $f$ is differentiable at $x=0$ and $f(0) = 0$, we know that for some $K$ and $\delta>0$ we have $|f(x)| = |f(x)-f(0)| \le K |x-0| = K|x|$ for any $x \in [0,\delta)$.
Hence we have $|g(x)| \le |x^{-{3 \over 2} } | K |x|= K {1\over \sqrt{|x|}}$ for all $x \in (0,\delta)$.
Furthermore, since $x \mapsto x^{-{3 \over 2} }$ is decreasing on $(0,1]$, we have $0\le |g(x)| \le ({\delta \over 2})^{-{3 \over 2}} |f(x)|$ for all $x \ge {\delta \over 2}$.
Combining, we see that
\begin{eqnarray}
\int |g| &\le& \int_{[0,{\delta \over 2})}|g| + \int_{[{\delta \over 2},1}|g| \\
&\le& \int_0^{\delta \over 2}K {1\over \sqrt{x}} dx + ({\delta \over 2})^{-{3 \over 2}} \int_{[{\delta \over 2},1} |f| \\
&=& \sqrt{\delta \over 2}K+{\delta \over 2}^{-{3 \over 2}} \int |f|
\end{eqnarray}
and so $g$ is integrable.
Since you're struggling with the concepts, first you have to show that each function $g_n(x)=2^{-n}f(x-r_n)$ is ($\mathcal{L},\mathcal{B}_{\mathbb{R}}$) measurable (the inverse image of a Borel set is in the Lebesgue $\sigma$-algebra). It suffices to check this for a set $(a,\infty), a\in\mathbb{R}$.
From there, you need to apply a limit theorem to the sum $s_n(x)=\sum_1^n g_n(x)$ (if two functions are measurable, their finite sum is measurable) that says that the limit of measurable functions is again measurable.
As a hint for the next part, verifying directly from the definition of the integral is difficult. I would say compute the integral $\int_{\mathbb{R}} g_n d\lambda$ first (just using basic calculus), and then think about applying one of the convergence theorems for Lebesgue integrals.
Best Answer
You have that $\mathbb{Q}$ is countable, so you can write $\mathbb{Q}=\bigcup_{n\geq 0}\{x_n\}.$ As a countable union of borelians sets, $\mathbb{Q}$ is a borelian set and so $\mathbb{1}_\mathbb{Q}$ is mesurable : by definition $$\int_\mathbb{R}\mathbb{1}_\mathbb{Q}\,d\mu=\mu(\mathbb{Q})=\sum_{n\geq 0}\mu(\{x_n\})=\sum_{n\geq0}0=0.$$