First we show that $g_1:=\sup_k f_k$ and $g_2 :=\inf_k f_k$ are measurable. Note that
$$g_1^{-1}(a,\infty] = \bigcup_k f_k^{-1}(a,\infty]
\quad\text{and}\quad
g_2^{-1}[-\infty,a) = \bigcup_k f_k^{-1}[-\infty,a)
$$
are measurable. Therefore the countable supremum and infimum of measurable functions are measurable. Consequently
$$
\limsup_k f_k = \inf_k(\sup_{j\geq k}f_j)
\quad\text{and}\quad
\limsup_k f_k = \sup_k(\inf_{j\geq k}f_j)
$$
are measurable functions.
Jumping in only because I find the terminology irritating.
A. Content zero vs measure zero.
Most authors use "content zero" to refer to the 19th century measure theory of Cantor-Peano-Jordan and "measure zero" to refer to the Lebesgue measure, dating from the first years of the 20th century.
You don't have to know much to connect them: a set has content zero if and only if the closure of the set has measure zero.
Nobody now or in the 19th century would have said (mainly because it is way too weak)
Bounded and set of discontinuities is content zero $\implies$ Riemann
integrable.
They knew back then exactly how to express this using the notion of content. We nowadays formulate this using "measure zero" since it is rather clumsy to state it using content.
Bounded and set of discontinuities is measure zero $\iff$ Riemann
integrable.
This is usually called the Lebesgue criterion for Riemann integrability. It is, however, just a restatement of an earlier criterion due to Hankel (1870), Volterra (1875) and Ascoli (1881). Lebesgue gets credit only for restating it using his measure.
B. Jump continuous vs regulated
The OP uses "jump continuous" to refer to a function whose only discontinuities are jump discontinuities. Yeech!
That is not an interesting class of functions really, especially since there is a larger class of functions of considerable importance. This strange usage suggests that removable discontinuities are not allowed?
Definition. A function is said to be regulated if it has one-sided limits at each point.
Properties:
- Every regulated function has only countably many discontinuities.
- All discontinuities are either removable or jump discontinuities.
- Step functions are regulated.
- A function is regulated if and only if it is the uniform limit of a sequence of step functions.
- All regulated functions are Riemann integrable.
C. Connections you ask?
A countable set of real numbers has measure zero. It need not have content zero. So regulated functions are Riemann integrable because all regulated functions are bounded and the set of discontinuities is countable and therefore has measure zero [not content zero please!].
Alternatively all regulated functions are uniform limits of step functions. But step functions are Riemann integrable and uniform limits of Riemann integrable functions are also Riemann integrable.
Best Answer
Let $f:[a,b]\rightarrow\mathbb{R}$ be a function with countable discontinuities, and let $c\in\mathbb{R}$. We must prove that $f^{-1}(-\infty,c)$ is measurable.
Let $A=\text{int}f^{-1}(-\infty,c)$ (the interior is taken in $[a,b]$) and $D=f^{-1}(-\infty,c)\setminus A$. Since $A$ is open, it is measurable, and since $f^{-1}(-\infty,c)=D\cup A$, it's only left for us to prove that $D$ is measurable. But notice that if $x\in D$, then $x$ is not an interior point of $f^{-1}(-\infty,c)$, that is, for every $r>0$, there exists $y$ such that $|y-x|<r$ but $f(y)\geq c$. Since $f(x)<c$, that means that $x$ is a discontinuity point of $f$. Therefore, $D$ is at most countable, hence measurable.
That argument can be used for any $f:E\subseteq\mathbb{R}^n\rightarrow\mathbb{R}$ which is Lebesgue-measurable, where $E$ is a Lebesgue-measurable subset of $\mathbb{R}^n$, and for which the set of discontinuities of $f$ has zero Lebesgue-measure.
(I'm using the following definition: a function $f:E\rightarrow\mathbb{R}$ is (Borel-)measurable iff for every $a\in\mathbb{R}$, $f^{-1}(-\infty,a)$ is a (Borel-)measurable set. It can be easily shown that any function which satisfies that condition also satisfies the following: for any Borel-measurable subset $A$ of $\mathbb{R}$, $f^{-1}(A)$ is a (Borel-)measurable subset of $E$)