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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 $f_n$.

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  • $\begingroup$ What measure do you use? $\endgroup$ – Hetebrij May 16 '16 at 9:43
  • $\begingroup$ @Hetebrij . My apologies. I edited the question with the missing information $\endgroup$ – user860374 May 16 '16 at 9:45
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    $\begingroup$ But isn't $\int f_n \text{d} \lambda = \infty$ for all $n \in \mathbb{N}$? $\endgroup$ – Hetebrij May 16 '16 at 9:50
  • $\begingroup$ @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. :) $\endgroup$ – user860374 May 16 '16 at 9:52
  • $\begingroup$ 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 $\lim\int f_n d\lambda=\infty$. $\endgroup$ – BigbearZzz May 16 '16 at 9:53
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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 $n \to \infty$.

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  • $\begingroup$ why so complicated? You could just say $$\int f_n d\lambda = \frac{1}{n}\cdot \lambda([n,\infty))$$ $\endgroup$ – user159517 May 16 '16 at 9:59
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    $\begingroup$ @user159517 Because DJS didn't know how to determine integrals of functions w.r.t. a measure, yet know when to switch integrals and limits, so I assumed it might be easier to start with integrals over set with finite measure. $\endgroup$ – Hetebrij May 16 '16 at 18:59

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