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I am supposed to give an example of a sequence of Riemann integrable functions $f_{n}$ and $f$, all defined on $[0,1]$ such that $\lim f_{n}(x)=f(x)$ but $\lim \int_{0}^{1} f_{n} (x)dx\neq\int_{0}^{1}f(x)dx$.

What I can think of is $f_{n}(x)=\max\{n-n^2|x-\frac{1}{n}|,0\}$. The graphs actually converges to a "spike" but the integral for each $f_{n}$ is always $1$ and so I have $\lim \int_{0}^{1} f_{n} (x)dx\neq\int_{0}^{1}f(x)dx$.

However, I am not sure if $\lim f_{n}(x)=f(x)$ is true at all $x$ in $[0,1]$ for my example because the vertex $(\frac{1}{n},n)$ goes to infinity.

LanaDR
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2 Answers2

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Your example, is correct: here's why.

It's better to write the above in a piecewise manner. Note that $f_n = n-n^2|x-\frac 1n|$ when $n > n^2|x-\frac 1n|$, or when $|x - \frac 1n| < \frac 1n$, or when $0 < x < \frac 2n$. Otherwise, it is $0$.

Clearly, the integral of each $f_n$ is $1$. But what is the limit function? That is, what is $\lim_{n\to \infty}f_n(x)$ for each $x$? Well, note that if $x=0$, then the limit is zero, since $f_n(0)=0$ for all $n$. If $x > 0$, then for some large enough $n$, $x > \frac 2n$, so $f_n(x) = 0$ beyond that point, and hence in the limit. Therefore, the limit function is $0$, and has integral zero, which means your example is correct.

Sarvesh Ravichandran Iyer
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Also we can consider easy example

$f_n(x)=n \Bbb 1_{(0, \frac {1}{n})}$ and $f(x)=0$

Then $\lim f_n(x)=f(x)$

And $\int f_n(x)=1$ for all $n\in \mathbb N$ and $\int f(x)=0$. Hence $\int f_n(x)\not \to \int f(x)$.

Mayuresh L
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