I am trying to come up with a proof of the following proposition:
Let $I\subset\mathbb{R}^n$ be an interval and let $(s_k)$ be a succession of staircase functions defined in $I$, increasing almost everywhere (a.e.) in $I$ such that $(\int_I s_k)$ is limited. Then exists an upper limit function $f$ such that $s_k\nearrow f$ a.e.
I've tried multiple approaches but have failed. I have a very clear notion of why this has to be true (if the set $D=\{x:\lim_{k\to\infty}s_k(x)=\infty\}$ isn't negligible then the integrals aren't bounded).
Any suggestions?
By increasing a.e. do you mean that, for almost every $x \in I$, $s_k(x) \le s_{k+1}(x)$ for all $k$? If so, for all such $x$, the sequence $\{s_k(x)\}$ converges, but possibly to $\infty$. Let $f(x) = \lim_k s_k(x)$ so that $f$ is defined, but possibly infinite, almost everywhere. If $s_1 \ge 0$ apply Fatou's lemma directly: $$ \int_I f = \int_I \lim_k s_k \le \liminf_k \int_I s_k < \infty. $$ Since $0 \le s_1 \le f$ and $f$ has finite integral, $f$ is finite almost everywhere and is the limit function you want.
If $s_1$ (and thus $f$) is not necessary nonnegative, apply a similar argument to the step functions $t_k = s_k - s_1$.