Suppose we have a real function $f$ bounded continuous on $[0,1]$.

We know that

$$ \frac 1 n \sum_{i=1}^n f(x_i) \to \int_0^1 f(x) \,dx$$

for $x_i \in [(i-1)/n, i/n]$, as $n\to \infty$.

Now suppose we have a sequence $f_n$ of bounded continuous functions on $[0,1]$ converging pointwise to $f$.

Is it true that

$$ \frac 1 n \sum_{i=1}^n f_n(x_i) \to \int_0^1 f(x) \,dx$$

  • $\begingroup$ $f$ need not be Riemann-integrable. $\endgroup$ – Chappers Jun 20 '18 at 22:45
  • $\begingroup$ So are you saying that it is true and that $f$ need not to be Riemann-integrable ? Can you provide a proof ? $\endgroup$ – W. Volante Jun 20 '18 at 22:47
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    $\begingroup$ math.stackexchange.com/q/108619/221811 , for example. $\endgroup$ – Chappers Jun 20 '18 at 22:53
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    $\begingroup$ What do you mean ? It is clearly stated that the $f_n$ and $f$ are continuous. $\endgroup$ – W. Volante Jun 20 '18 at 23:53
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    $\begingroup$ In such a case, by considering $f_n-f$, your question can be stated as: is it true that any sequence $\{g_n\}_{n\geq 1}$ of continuous functions on $[0,1]$, such that $g_n(x)$ is pointwise convergent to $0$ for any $x\in[0,1]$, is such that $$\lim_{n\to +\infty}\int_{0}^{1}g_n(x)\,dx = 0$$ ? Unluckily, if the hypothesis of the dominated convergence theorem are not met, the answer still is not necessarily. On the other hand, if the $g_n$ are equi-bounded, the outcome is positive. $\endgroup$ – Jack D'Aurizio Jun 21 '18 at 0:25

Consider the sequence of continuous functions $\{f_n\}_{n=1}^\infty$ defined on $[0,1]$ that arises by setting $f_1(x)=x$, and for $n=2,3,\ldots$ \begin{align}f_n(x) := \begin{cases} n^2x , \; \text{ if } x \in \left[0 , \,\frac{1}{n}\right], \\ 2n - n^2x , \; \text{ if } x \in \left[\frac{1}{n} , \,\frac{2}{n}\right], \\ 0, \; \text{ if } x \in \left[\frac{2}{n}, \,1\right].\end{cases} \end{align}

So for $n=1,2, \ldots$ we have $|f_n(x)| \leq n$, for all $x \in [0,1]$ (each function in the sequence is bounded on $[0,1]$). Moreover, $\lim\limits_{n\to \infty} f_n(x)=0$ and so the pointwise limit is the bounded and continuous function $f(x):=0$ $(x \in [0,1])$. For $n=1,2,...$ let $S_n := \big\{\frac{i}{n} \in [0,1] : i \in \mathbb{N}\big\}$ and for $i=1,\ldots,n$ we allow $x_i \in S_n$ to denote the number $\frac{i}{n}$. Then we have \begin{equation}\frac{1}{n} \sum_{x_i \in S_n} f_n(x_i) = 1 \;\;\; (n=1,2,\ldots) \end{equation} but $\frac{1}{n} \underset{x_i \in S_n}{\sum} f(x_i)=0$ for all $n \in \mathbb{N}$ (notice that for $n=2,3,\ldots$ we have $f_n(x_{n-1})=n$ ). In other words, $ \lim\limits_{n \to \infty} \frac{1}{n} \overset{n}{\underset{i =1}\sum} f_n\left(\frac{i}{n}\right) = 1$ (limit of a constant sequence) but $ \int_0^1 f(x) \,dx=0$.

Since this is a counterexample, the statement is not true...

As a bonus:

enter image description here

  • $\begingroup$ $f_1(x)=1$ is another good choice, call it a matter of preference (value of the integral or definition of a function in the sequence). $\endgroup$ – Matt A Pelto Jun 21 '18 at 4:09
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    $\begingroup$ Thank you. If $f_n$ is uniformly bounded, do you think the statement becomes true ? $\endgroup$ – W. Volante Jun 21 '18 at 9:13
  • $\begingroup$ Yes it does. For a proof, see Arzelà's Dominated Convergence Theorem for the Riemann Integral here: sites.math.washington.edu/~morrow/335_15/dominated.pdf $\endgroup$ – Matt A Pelto Jun 21 '18 at 15:40

Let $f_n(x) = n^2x^n(1-x).$ Then $f_n\to 0$ pointwise everywhere on $[0,1].$ But

$$\frac{1}{n}\sum_{i=1}^{n}f_n(i/n) > \frac{1}{n}f_n((n-1)/n) = \frac{1}{n}n^2(1-1/n)^n\frac{1}{n} \to \frac{1}{e} \ne \int_0^1 0\,dx.$$


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