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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 $1 \leqq q<n, 0 \leqq p \leqq q$, 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!

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