find $\{f_n \}$ a sequence of continuous functions on $\mathbb{R}$ such that $0 \leq f_n(x) \leq 1$ and $\lim_{n \rightarrow \infty} \int_0^1 f_n(x)dx =0$, but $f_n(x)$ does not converge anywhere on $[0,1]$.

Everytime I am thinking with a sequence, I come with one satisfies some of the conditions but violate one condition. For example, $f_n(x)=x^n$ is continuous sequence, and $f_n \rightarrow 0$ so limit the integral is zero, but it is pointwise convergent. , I tried to create a function similar to Dirchlet function, as they not converge pointwise, but they converge on part of the domain and they are not continuous.

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    $\begingroup$ You start with a conditional, “If”, but there is never a conclusion (“then”). The sentence is not grammatical and you should really say what you mean. I think “Does there exist...” rather than “If”... $\endgroup$ Mar 23, 2021 at 23:41
  • $\begingroup$ I updated the question! $\endgroup$ Mar 23, 2021 at 23:56

1 Answer 1


Rather than give formulas, I will describe the graphs of the functions, which are piecewise linear. Think of a “discretely moving wave”, where after the wave is done moving, we switch to a “thinner” wave.

Let $f_1$ go from $(0,1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Let $f_2$ go from $(0,0)$ to $\frac{1}{2},1)$ to $(1,0)$. Let $f_3$ go from $(0,0)$ to $\frac{1}{2},0)$ to $(1,1)$.

Then let $f_4$ go from $(0,1)$ to $(\frac{1}{4},0)$ to $(1,0)$. Let $f_5$ go from $(0,0)$ to $(\frac{1}{4},1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Then $f_6$ from $(0,0)$ to $(\frac{1}{4},0)$ to $(\frac{1}{2},1)$ to $(\frac{3}{4},0)$ to $(1,0)$. Then $f_7$ go from $(0,0)$ to $(\frac{1}{2},0)$ to $(\frac{3}{4},1)$ to $(1,0)$. Then $f_8$ from $(0,0)$ to $(\frac{3}{4},0)$ to $(1,1)$.

Next do the same thing with the partition $0\lt \frac{1}{8}\lt\frac{1}{4}\lt\frac{3}{8}\lt \frac{1}{2}\lt \frac{5}{8}\lt\frac{3}{4}\lt\frac{7}{8}\lt 1$. Then partitioning $[0,1]$ into $16$ equal subintervals, etc.

Each new batch of functions has a smaller integral, all positive, but converging to $0$. The functions are all continuous. And $f_n(x)$ does not converge for any $x$ because you can always find arbitrarily large values of $n$ where $f_n(x)=0$ and values where $f_n(x)$ is very close to $1$ (by approximating $x$ with a rational with denominator a power of $2$).

  • $\begingroup$ I am curious to ask if these functions have a certain pattern or compact formula!! $\endgroup$ Mar 23, 2021 at 23:58
  • $\begingroup$ It is really a clever example! $\endgroup$ Mar 23, 2021 at 23:58
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    $\begingroup$ @math_for_ever: I’m sure we could come up with a relatively succint formula in terms of the nearest power of $2$ or something, but I don’t think there’s much point in expending the effort... $\endgroup$ Mar 23, 2021 at 23:59
  • $\begingroup$ you are right! Does this sequence has a name? $\endgroup$ Mar 24, 2021 at 0:11
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    $\begingroup$ @fourelements the subsequence consisting of the very first wave of each size converges pointwise to $\chi_{\{x=0\}}$. $\endgroup$
    – Alan
    Jul 11, 2021 at 23:53

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