# Lebesgue integral - no dominating integrable function of $(f_n)$

Let $\lambda$ be the Lebesgue-measure on $\Omega =[0,1]$. Given a sequence of non-negative measurable functions $$f_n:\Omega\to\Bbb R: x \mapsto ne^{-nx},$$ how can I show that $f_n$ converges $\lambda$-almost everywhere to a measurable function $f$, but $$\int f d\lambda \neq \lim_{n\to\infty} \int f_nd\lambda$$

The theorem of dominated convergence seems to fail here..

Any hints for the proof?

Edit: Ok what I have done so far: $(f_n)$ converges almost everywhere to $f(x)=0$. Obviously $f_n(0)=n$ for all $n \in \Bbb N_{\gt 0}$. Hence, $$\int f d\lambda =0 \neq 1 = \lim_{n\to\infty} \int f_nd\lambda$$

Now, how can I show that there is no integrable function that dominates $(f_n)$ without utilizing the theorem of dominated convergence?

• Did you try to compute the limit of $f_n$? – Guy May 20 '16 at 10:16
• I think it is $\lim_{n\to\infty} f_n = 0$ ? So the measurable funcion $(f_n)$ converges to is $f(x)=0$? – Tesla May 20 '16 at 10:21
• OK, next you need to compute $\int f_n\;d\lambda$. – GEdgar May 20 '16 at 12:41

For $x\in(0,1]$ we have \begin{align} \lim_{n\to\infty} ne^{-nx}&=e^{\lim_{n\to\infty}\left(\log n-nx\right)}=0, \end{align} and for $x=0$ $$\lim_{n\to\infty}ne^{-n\cdot 0}=\infty.$$ Therefore $f_n\to 0$ a.e. and hence $$\int_{[0,1]}\lim_{n\to\infty} f_n\ \mathsf d\lambda=0.$$ However, $$\int_{[0,1]}f_n\ \mathsf d\lambda = \int_0^1 ne^{-nx}\ \mathsf dx=-e^{-nx}|_0^1 =1-e^{-nx}.$$ For $x\in(0,1]$ we have $$\lim_{n\to\infty} (1-e^{-nx}) = 1,$$ and thus $$\lim_{n\to\infty}\int_{[0,1]}f_n\ \mathsf d\lambda = 1.$$
• @Sigma This shows that "$f_n$ converges $\lambda$-almost everywhere to a measurable function $f$, but $$\int_{[0,1]}f\ \mathsf d\lambda \ne \lim_{n\to\infty}\int_{[0,1]}f_n\ \mathsf d\lambda,"$$ precisely the question asked. No need to invoke dominated convergence here (and in fact, such an argument would be incorrect, since its converse is not true). – Math1000 May 20 '16 at 19:23
• The question is why I cannot find a dominating function $g$ such that $|f_n(x)| \le g(x)$ for all $x$ ? Of course, our given solution implies that there cannot be any but the question is why there cannot be any. – Tesla May 21 '16 at 6:40
• There is no dominating function because if there would be some, the integral of the limit would be the limit of the integral. (Argument by contradiction.) Alternatively, you could also try to compute the smallest function that lies above all $f_n$ and show that it is not integrable. – Dirk May 21 '16 at 13:26
• If you use your argument by contradiction, you in fact did utilize the dominated convergence theorem which shall not be utilized to show that there can't exist a dominating function $g$. – Tesla May 21 '16 at 14:04