$f_n = (\frac{1}{n})\chi_{[n, +\infty)}$. Find $\lim \int f_n d\lambda$.

Let $X = \mathbb R$, $\textbf{X} = \textbf{B}$ and $\lambda$ the Lebesgue measure on $\textbf{X}$.

I have the following: $f_n = (\frac{1}{n})\chi_{[n, +\infty)}$.

I need to find the following: $\displaystyle \lim \int f_n d\lambda$.

I honestly have no idea how to find this, since $(f_n)$ is not a monotone increasing sequence of functions I cannot say that $$\int f d\lambda = \lim \int f_n d\lambda$$ where $f=0$, the uniform limit of $f_n$.

• What measure do you use? – Hetebrij May 16 '16 at 9:43
• @Hetebrij . My apologies. I edited the question with the missing information – user860374 May 16 '16 at 9:45
• But isn't $\int f_n \text{d} \lambda = \infty$ for all $n \in \mathbb{N}$? – Hetebrij May 16 '16 at 9:50
• @Hetebrij , how do I show that though? Could you please show me? :) I'm not sure how to determine integrals of functions wrt a measure yet, since this is the first example I'm working with. :) – user860374 May 16 '16 at 9:52
• I'd suppose that this is a counter example to $\lim\int f_n d\lambda = \int\lim f_n d\lambda$. It's not hard to see that $\int\lim f_n d\lambda=0$ while $\lim\int f_n d\lambda=\infty$. – BigbearZzz May 16 '16 at 9:53

For a set $A$ of finite measure, we have $\int a\chi_{A} \text{d} \lambda = a\lambda(A)$ for all $a \in \mathbb{R}$.
We note that for $f_n$, we have $g_{n,m} = \frac{1}{n} \chi_{[n,m]}$ for $m>n$ converges pointwise to $f_n$ and is increasing to $f_n$, so in this case we may interchange limit and integration. So we have $$\int f_n \text{d} \lambda = \lim_{m \to \infty} \int g_{n,m} \text{d} \lambda = \lim_{m \to \infty} \frac{m-n}{n} = \infty.$$ So for all $n \in \mathbb{N}$, we have $\int f_n \text{d} \lambda = \infty$, and so is their limit as $n \to \infty$.
• why so complicated? You could just say $$\int f_n d\lambda = \frac{1}{n}\cdot \lambda([n,\infty))$$ – user159517 May 16 '16 at 9:59