# Example of sequence of continuous function which converge pointwise a.e. but does not converge uniformly on any subinterval

I want to construct a sequence of continuous function which converge pointwise a.e. but does not converge uniformly on any subinterval. from David Mitra's answer (https://math.stackexchange.com/q/405594), I found an answer in the book Counterexamples in Analysis, Gelbaum and Olmsted , as below:

A more interesting example is given by use of the function $$f$$ (cf. Example 15, Chapter 2) defined: $$f(x) \equiv\left\{\begin{array}{cc}\frac{1}{q} & \text { if } \quad x=\frac{p}{q} \text { in lowest terms, where } p \text { and } q \text { are integers } \\ \text { and } q>0 .\end{array}\right.$$ For an arbitrary positive integer $$n$$, define $$f_{n}(x)$$ as follows: According to each point $$\left(\frac{p}{q}, \frac{1}{q}\right)$$, where $$1 \leqq q, in each interval of the form $$\left(\frac{p}{q}-\frac{1}{2 n^{2}}, \frac{p}{q}\right)$$ define $$f_{n}(x) \equiv \min \left(\frac{1}{n}, \frac{1}{q}+2 n^{2}\left(x-\frac{p}{q}\right)\right)$$ in each interval of the form $$\left(\frac{p}{q}, \frac{p}{q}+\frac{1}{2 n^{2}}\right)$$ define $$f_{n}(x) \equiv \max \left(\frac{1}{n}, \frac{1}{q}-2 n^{2}\left(x-\frac{p}{q}\right)\right) ;$$ and at every point $$x$$ of $$[0,1]$$ at which $$f_{n}(x)$$ has not already been defined, let $$f_{n}(x) \equiv 1 / n$$. Outside $$[0,1] f_{n}(x)$$ is defined so as to be periodic with period one. The graph of $$f_{n}(x)$$, then, consists of an infinite polygonal arc made up of segments that either lie along the horizontal line $$y=1 / n$$ or rise with slope $$\pm 2 n^{2}$$ to the isolated points of the graph of $$f$$. (Cf. Fig. 2.) As $$n$$ increases, these "spikes" sharpen, and the base approaches the $$x$$ axis. As a consequence, for each $$x \in Q$$ and $$n=1,2, \cdots$$, and $$f_{n}(x) \geqq f_{n+1}(x),$$ $$\lim _{n \rightarrow+\infty} f_{n}(x)=f(x),$$ as defined above. Each function $$f_{n}$$ is everywhere continuous, but the limit function $$f$$ is discontinuous on the dense set $$Q$$ of rational numbers.

My question is: If we define $$f_n(x)≡0$$ instead of 1/n at every point x of [0,1] at which $$f_n(x)$$ has not already been defined,(And do not use 1/n in the definition of $$f_n(x)$$ in the "spike", let $$f_{n}(x) \equiv \frac{1}{q}\pm 2 n^{2}\left(x-\frac{p}{q}\right)$$), it seems the sequence still has the property we want. Am I wrong? Or why we need a "$$\frac{1}{n}$$"? Thanks!