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.
It is worth mentioning that the origin of the question lies in queue dynamics of dynamic flows (also called flows over time) with the so-called Vickrey's deterministic fluid queuing.
More specifically, I found an answer in the paper [1, Section 2.2] of Cominetti, Correa und Larré.
I worked out the relevant proofs in the following.
It has proven beneficial to first analyze the uniqueness of the queue function $q(t):= \int_0^t f - g \,\mathrm{d}\lambda$.
Claim 1. Given a function $g$ fulfilling (1) and (2) a.e., the queue function is of the form $$q(t) = \max_{u\in[0,t]} \int_u^t f - \nu \, \mathrm d\lambda.$$
Proof.
We first show, that $q$ is never negative.
Assume the contrary and let $t>0$ fulfill $q(t) < 0$.
Choose $u^* := \max\{u\leq t \,\vert\, q(u) = 0 \}$ to be the latest time before $t$ at which the queue was zero.
By the continuity of $q$ we know that $q$ is strictly negative on $(u^*, t]$.
Thus using property (2) we get $$q(t) = \int_0^t f - g \,\mathrm d\lambda = \int_0^{u^*} f - g \,\mathrm d\lambda + \int_{u^*}^t f - f \,\mathrm d\lambda = q(u^{*})=0,$$
a contradiction.
Now, we focus on the statement of the claim. Let $t\in \mathbb R$ and let us define $u^* := \max\{u\leq t \,\vert\, q(u) = 0 \}$ again to be the latest time before $t$ at which the queue is empty.
Then by definition we have $q(u) > 0$ for all $u\in(u^*, t]$ and hence (2) implies $g(u) = \nu$ for almost all $u\in(u^*, t]$.
This shows $$q(t) = \int_0^{u^*} f - g \,\mathrm d\lambda + \int_{u^*}^t f - g \,\mathrm{d}\lambda = q(u^*) - \int_{u^*}^t f - \nu \,\mathrm{d}\lambda = \int_{u^*}^t f - \nu \,\mathrm{d}\lambda.$$
It remains to show, that $\int_u^t f - \nu \,\mathrm d\lambda \leq q(t)$ holds for any other $u\in[0,t]$.
If $u\leq u^*$, then (1) implies $$\int_u^t f - \nu \,\mathrm d\lambda\leq \int_u^t f - g\,\mathrm d\lambda = q(t) - q(u) \leq q(t).$$
If $u > u^*$, we use (2) and get $$\int_u^t f - \nu \,\mathrm d\lambda = q(t) - \int_{u^*}^{u}f - \nu \,\mathrm d\lambda - q(u^*) = q(t) - q(u) \leq q(t). $$
$\square$
Note: We haven't used that $\nu$ is constant. Instead it could be any locally integrable function.
Claim 2. The function $q(t) := \max_{u\in[0,t]} \int_u^t f - \nu \, \mathrm d\lambda$ is almost everywhere differentiable with $$
q'(t) = \begin{cases}
f(t) - \nu, &\text{if } q(t) > 0,\\
0, &\text{otherwise}.
\end{cases}
$$
Moreover, we have $f(t) \leq \nu(t)$ for almost all $t$ with $q(t)=0$.
I refer to this answer for a worked out proof.
The "Moreover"-part is proven here.
Note: We could replace $\nu$ by any non-negative locally integrable function.
Final Claim. Given a locally integrable function $f:\mathbb R_{\geq 0}\rightarrow \mathbb R_{\geq0}$, there exists a locally integrable $g: \mathbb R_{\geq 0}\rightarrow \mathbb R_{\geq0}$ fulfilling (1) and (2) a.e..
Moreover, $g$ is unique up to a null set.
Proof.
We first define $q(t) := \max_{u\in[0,t]} \int_u^t f - \nu \,\mathrm d \lambda$.
We define $g(t) := f(t) - q'(t)$ wherever $q'(t)$ is defined, and $g(t):=0$ elsewhere.
Then, we have $q(t) = \int_0^t f - g \,\mathrm d\lambda$ and $g$ automatically fulfills (2) a.e. by Claim 2.
It remains to show, that $g(t)\leq \nu$ almost everywhere. For $q(t) > 0$, this is given by (2) already.
For $q(t)=0$ this follows from Claim 2.
Now assume there is another function $h$ fulfilling (1) and (2) almost everywhere. By Claim 1 we have $$\int_0^t f - g \,\mathrm d\lambda = \max_{u\in[0,t]} \int_u^t f -\nu \,\mathrm d\lambda = \int_0^t f -h \,\mathrm d\lambda.$$
for any $t\geq 0$. Hence $g$ and $h$ coincide almost everywhere.
$\square$
Note: We haven't used that $\nu$ is constant. Instead it could be any locally integrable function.
References
[1] "Dynamic Equilibria in Fluid Queuing Networks" by Roberto Cominetti, José R. Correa, Omar Larré. Available at arXiv:1401.6914 [math.OC].
Best Answer
Your idea about the measurability is correct.
Now i pressume that you mean with $s>0$ that $u(x)=\frac{1}{x^s}$ when $x \in (0,1]$
For $0<s<1$ the function is integrable and you can also compute its integral.
For $s \geq 1$ we have that $$\int_0^1 u(x)dx \geq \sum_{k=1}^{\infty}\int_{\frac{1}{2n+2}}^{\frac{1}{2n+1}}\frac{1}{x^s}dx$$ $$ \geq \sum_{k=1}^{\infty} (2n+1)^s\frac{1}{(2n+1)(2n+2)}$$ $$= \sum_{k=1}^{\infty}\frac{(2n+1)^{s-1}}{2n+2}$$ $$\geq \sum_{k=1}^{\infty} \frac{(2n)^{s-1}}{4n}=+\infty$$ since $s-1 \geq 0$