We define $f$ to be lebesgue integrable on set $E$ if
$P: E = \sqcup_{k=1}^NE_k,$$ $$\inf_PU(P, f) = inf_P\sum_{k=1}^N\sup_{E_k} f(x)\mu(E_k) = sup_P\sum_{k=1}^N\inf_{E_k}f(x)\mu(E_k)= sup_PL(P, f),$
where $\mu$ is a Lebesgue measure.
We don't assume $f$ is measurable. In all subsequent theorems, we explicitly emphasize "If $f$ is measurable and ...". But does exist a function that is not measurable, but is Lebesgue integrable by definition I wrote?
I will try to answer your question for indicator functions $\chi_S$. This may give some indication how the question could be answered for simple functions, and perhaps arbitrary functions.
For any given partition $E = \bigsqcup_{k=1}^n E_k$, notice $\sup_{x\in E_k} \chi_S(x)$ is $1$ if $E_k$ hits $S$ and $0$ otherwise.
Let $A$ be the union of all the $E_k$'s that hit $S$ and $B$ be the union of the other $E_k$'s.
Let $Q$ be the partition $E = A \sqcup B$. Then $U(P, \chi_S) = U(Q, \chi_S) = \mu(A)$.
We see that $\inf_P U(P, \chi_S)$ is simply the inf of the measures of Lebesgue supersets of $S$.
Likewise $\sup_P L(P, \chi_S)$ is the sup of the measures of the Lebesgue subsets of $S$.
$S$ is measurable iff these two numbers agree.