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.
In addition to the previous advice, note that the function you gave does not "approximate" $f$. An approximating sequence $\{s_n\}$ for $f$ (which $f$ would have iff $f$ were measurable provided the measure space for the domain of $f$ is complete) would need to be within a distance of $\epsilon$ from $f$ for any given $\epsilon >0$. However, the function $s$ you gave as an "approximation" cannot approximate $f$, in the sense that the sequence $\{s_n\}$ with $s_n=s$ $ \forall n \in \mathbb{N}$ gets no closer than $1$ from $\chi_{[0,1]}$.
Again, it is important to note that a function $f$ on a complete measure space $X$ is measurable if and only if $f$ is the pointwise limit of some sequence of simple functions -or, trivially, is a simple function itself. Consequently, any sequence of "simple" functions approximating a nonmeasurable function must contain a "simple" function with the characteristic function of a nonmeasurable set as part of its construction. In a sense, this is why you must include the assumption that $f$ is measurable in your definition for the Lebesgue integral.
To wit, recall that the integral of a characteristic function is the measure of the pullback set in your domain; in the case of your $f$, since $f$ is characteristic, the integral, were it to be defined, is the measure of the pullback, $$\int f d\mu = \mu\{f^{-1}\{1\}\}=\mu\{[0,1]\},$$
but you have not defined the measure for $[0,1]$, which is not even in your $\sigma$-algebra; nor can we infer the measure of $[0,1]$ from the definition you gave for your measure space, as your collection of measurable subsets is already a closed $\sigma$-algebra that is $\sigma$-finite under $\mu$ (hence, your measure space cannot even be extended, in the usual Caratheodory way, to include $[0,1]$ with an accompanying well-defined measurement).
As a curiosity tangential to your question, it is possible for nonmeasurable functions to arise from limits of simple functions in a complete measure space, but such a collection of simple functions must be uncountable. For instance, with Lebesgue measure on $\mathbb{R}$, take $f=\chi_V$ to be the characteristic function of the (uncountable and nonmeasurable) Vitali set $V$ on $[0,1]$, and consider the (uncountable) collection of measurable functions $\{\chi_v\}, v\in V$. Then $\chi_V=\sup \{\chi_v\}_{v\in V}$. Were we to define an integral as you wish, then in this case, you may want to say that the integral $\int \chi_V = \sup \{ \int \chi_v \} = 0$. But, again, since $\chi_V$ is itself characteristic, we should then have $\mu(V)=0$, but $\mu(V)$ is not defined for $V$ under the Lebesgue measure.
To address your request for a resource, see Royden's Real Analysis, 4th ed., chapters 17 and 18 (particularly pp. 362-363 were helpful as a reference to me for this post).
Best Answer
Without measurability, that definition could be called the lower Lebesgue integral $\underline{\int} f \,d\lambda$. Unfortunately, it lacks desirable properties. For example $$ \underline{\int}(f+g)\;d\;\lambda = \underline{\int}f\;d\;\lambda + \underline{\int}g\;d\;\lambda $$ may easily fail.
If $E$ is a Bernstein set in $[0,1]$, let $f = \mathbf1_E$ be its indicator function, and let $g = \mathbf1_{[0,1]\setminus E}$. Then $$ \underline{\int}(f+g)\;d\;\lambda = 1,\\ \underline{\int}f\;d\;\lambda = 0,\\ \underline{\int}g\;d\;\lambda = 0. $$