In France, we learn that a function $f$ on an interval $I$ is said to be piecewise continuous if it is piecewise continuous on any segment included in $I$.
Therefore, the function defined on $(0,1]$ that takes the value $\frac1n$ on $\left(\frac{1}{n+1},\frac1n\right]$ for $n\geq 1$ is piecewise continuous.
However, the natural extension to $[0,1]$ is not piecewise continuous since the subdivision is then infinite.
It feels a bit odd. I had the impression that elsewhere, the convention is that a function on a bounded interval (whatever the type of the interval) is said piecewise continuous if there is a finite subdivision of this interval into open intervals and such that each restriction can be extended to a continuous function on the closure of these intervals. As for unbounded interval, the subdivision is locally finite.
With this convention, both function above are not piecewise continuous. Is there a reference for these conventions ?
Are you sure that the definition you have mentioned is complete? Since in this form as you have mentioned, your example is also not piecewise continuous by taking any segment containinf the jump discontinuity. That argument is used for infinite intervals onlym give it a look here: Piecewise continuity with countably infinite discontinuities