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A jump continuous function is defined as a function such that all the lateral limits of it points are well-defined. I think that the statement of the title is true, can you check my proof please?

We previously knows that a jump continuous function have, at most, a countable number of discontinuities, and this subset of discontinuities is a set of isolated points.

Then what define jump continuous functions is it continuity at any lateral neighborhood (if exists) of any (limit) point of it domain. Let define a left-neighborhood of some point $x$ as the interval $(x-\delta,x)$ for some $\delta>0$, and a right-neighborhood of $x$ as the interval $(x,x+\delta)$.

Suppose that $f$ have a discontinuity at some $g(x)$. I claim that exists lateral neighborhoods at the sides of $x$ such that $f\circ g$ is continuous. Proof: there are two cases:

  1. If $g$ is constant in some lateral neighborhood then the image of this neighborhood is a point, hence $f\circ g$ is continuous in this neighborhood.

  2. If $g$ is not constant at some lateral neighborhood then the image of this neighborhood is an interval. Because $f$ is jump continuous then for any point in that interval the function is continuous at it laterals, hence there is a lateral neighborhood of any point where the function is continuous, what imply that $f\circ g$ is continuous in this neighborhood.

By the other side if $g$ have a discontinuity at $x$ then $g$ is continuous at it lateral neighborhoods, hence, by the same reasons shown above, exists lateral neighborhoods of $x$ such that $f\circ g$ is continuous.

Hence for any point of $\mathrm{dom}(f)$ exists lateral neighborhoods where $f\circ g$ is continuous, hence $f\circ g$ is jump continuous.$\Box$.

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The statement is false, because you would need some sort of monotonicity for the limit of the composition to exist. As a counterexample, consider $$ f(x)=\begin{cases} x\sin(1/x) &\text{ if } x\neq 0\\ 0 &\text{ if } x=0, \end{cases} $$ and $$ g(x)=\begin{cases} 1 &\text{ if } x\geq 0\\ -1 &\text{ if } x<0, \end{cases} $$ $f$ is continuous, hence jump continuous, and $g$ is obviously jump continuous. The composition $g\circ f$ does not have a left lateral limit, nor a right one, at $x=0$.

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    Oh, I see... In first place I was searching some counter-example but I dont found one so I supposed that it was true... Sad. – 2017-02-08
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Hint: Consider $f(x) = x$ for $x> 0$ and $f(x) = -1$ for $x\le 0$. Is $1/f(x)$ jump continuous?

For a bounded example, consider $\sin(1/f(x))$.

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    Oh, right, the function must be bounded... – 2017-02-08
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    @Masacroso: see edit – 2017-02-08
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    $\sin(1/f)$ is not jump continuous, as it does not have a right limit at $x=0$. – 2017-02-08
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    @zugg: that is the point. $f$ is jump continuous, an $\sin(1/y)$ is continuous on $\mathbb R\setminus \{ 0 \}$. – 2017-02-08
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    I'm assuming the OP was asking about a composition $g\circ f$ where $g$ is jump continuous on $f(\mathbb R)$. If not, then your counter-example stands. – 2017-02-08
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    @zugg: you do notice that $f(\mathbb R) = \{ -1 \} \cup (0,\infty) \subseteq \mathbb R \setminus \{ 0 \}$, right? And a continuous function is obviously also jump continuous. – 2017-02-08
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    Sorry, yes. It fails to be jump continous only on $\overline{f(\mathbb R)}$. – 2017-02-08
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    @zugg: yes, but op hasn't asked for it. – 2017-02-08