# Example with almost every convergence where the dominated convergence theorem fails

So I ran into this exercise, and I want someone to check the accuracy of my answer.

Let $f_n(x) : \mathbb{R} \to \mathbb{R}, f_n(x)= n\mathcal{X}_{[0,\frac{2}{n}]} \forall$ n $\in \mathbb{N}$. Check whether the functions $f_n$ are $\lambda$-integrable and whether $f=\lim_{n \to \infty}{f_n}$ exists or not, pointwise. After that, check whether $\int{f_n}dλ \to \int{f}dλ$. What can you observe about the theorems of dominated convergence and the monotone convergence?

~Well, $\int{f_n}dλ = \int_{[0,\frac{2}{n}]}{n}dλ + 0 = 2$ so it is finite, hence $\lambda$-integrable. I'm not sure if it is ok to say that $f_n \to f=0 a.e$ since the only point that $f_n$ does not converge to $0$ is $0$, but $λ(\{0\})=0$.

Since as we saw $\int{f_n}dλ = \int_{[0,\frac{2}{n}]}{n}dλ + 0 = 2 \to 2$, but $\int{f}dλ=0$, the convergence stated,$\int{f_n}dλ \to \int{f}dλ$, is not true.

Dominated convergence cannot hold, since there is not a sufficient bound for $f_n$, and (unless I can't see it) $f_n$ is not an increasing sequence, so monotone convergence theorem cannot be used either.

## 1 Answer

Your thought are correct; maybe when we check that $f_n$ is integrable we should put the absolute value (the function is non-negative, but it shows that you didn't forget it).

You are right about pointwise convergence. The convergence is not monotonic (if $x\in (0,1)$ is fixed, then $f_n(x)$ increases then it becomes $0$ for $n$ large enough depending on $x$) and since $\sup_n|f_n|$ is not integrable there is no dominating integrable function.