# Discontinuity properties of $f_n$ carries over to the limit function $f$

Suppose that $$f_n:[a,b] \rightarrow \Bbb R$$ and $$f_n$$ converges uniformly to $$f$$. Which of the following discontinuity properties of the functions $$f_n$$ carries over to the limit function ?

• No discontinuities
• At most ten discontinuities
• At least ten discontinuities
• Uncountably many discontinuities
• Countably many discontinuities
• No jump discontinuities
• No oscillating discontinuities

My try :

For first bullet: If each $$f_n$$ is continuous and convergence is uniform, then by $$\varepsilon/3$$ argument, $$f$$ is continuous, which means $$f$$ has no discontinuities.

For fourth bullet: To disprove this , consider $$f_n(x)=\begin{cases} \frac{1}{n} & \text{if}\; x \in \Bbb Q \cap [0,1]\\\\0 &\text{otherwise}\end{cases}$$

Then $$f_n$$ is discontinuous everywhere on $$[0,1]$$ whereas the limit $$f=0$$ is continuous. Of course , the convergence is uniform

For fifth bullet: To disprove this, consider $$f_n(x)=\begin{cases} \frac{1}{n} & \text{if}\; 0

Here $$f_n$$ converges uniformly to $$f=0$$ and each $$f_n$$ has discontinuous at $$x=0$$ and $$x=\frac{1}{n}$$ but $$f$$ continuous on $$[0,1]$$.

Is my arguments correct ? Can I have a hint for others ?

• My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite. – zhw. Dec 3 '18 at 18:55
• The first bullet point is exactly about the case where $f_n$ has no discontinuities. – Ingix Dec 4 '18 at 8:35

Hint for second: This property also propagates from the $$f_n$$ to $$f$$. The proof is similar to the proof for the first bullet point: If $$f$$ has a discontinuity at some $$x_0$$, then the $$f_n$$ must also have a discontinuity at $$x_0$$ for all $$n>N$$ for some suitable $$N$$.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $$f$$ that does have countably many discontinuities (namely $$0$$). For a hint, prove and then use that if all $$f_n$$ are continuous at some point $$x_0$$, $$f$$ is also continuous at $$x_0$$.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $$f_n$$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.