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Consider, on the measure space of reals, the sequence $f_1,f_2,\dots$ with $f_n(x)=1$, if $n \leq x \leq n+1$, and zero otherwise. It seems that $\int f_n d\mu=1$ for all $n$. However, the claim is that $\lim_{n\to \infty}f_n=0$. This last limit I want to understand.

Do I prove it via the definition of a limit? Namely, $$ \forall \epsilon >0 \exists n_o\in \mathbb{N}\forall n\geq n_0 (|f_n|<\epsilon), $$ and try to find an appropriate $n_0=n_0(\epsilon)$? Is there any general way of understanding what the limit of functions is?

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  • $\begingroup$ Drawing graphs of $f_n$ for small values of $n$ may be useful. $\endgroup$
    – Cop 663
    Apr 1 '19 at 16:50
  • $\begingroup$ It's "humps" of unit area, being successively displaced to the right. I do not see, however, how in the limit as the humps are displaced to infinity, the function is zero. $\endgroup$
    – EEEB
    Apr 1 '19 at 16:52
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We say that $(f_n)_{n=1}^\infty$ converges to $f$ pointwise if for every $\varepsilon>0$ and every $x\in\mathbb R$, there exists an $N_{\varepsilon,x}>0$ such that for every $n\geq N_{\varepsilon,x}$, it is the case that $|f_n(x)-f(x)|<\varepsilon$.

We say that $(f_n)_{n=1}^\infty$ converges to $f$ uniformly if for every $\varepsilon>0$, there exists an $N_\varepsilon>0$ such that for every $x\in\mathbb R$ and every $n\geq N_\varepsilon$, it is the case that $|f_n(x)-f(x)|<\varepsilon$.

Let $f_n$ be as you prescribed and let $f(x)=0$ for every $x\in\mathbb R$.

To see that $(f_n)_{n=1}^\infty$ converges to $f$ pointwise, let $\varepsilon>0$, let $x\in\mathbb R$, and let $N_{\varepsilon,x}=x+1$. Then observe that for every $n\geq N_{\varepsilon,x}$, it is the case that $|f_n(x)-f(x)|=f_n(x)=0<\varepsilon$.

To see that $f_n$ does not converge to $f$ uniformly, let $\varepsilon=1/2$, let $N_{\varepsilon}>0$, and let $x_n\in[n,n+1]$. Then observe that for every $n\geq N_\varepsilon$, it is the case that $|f_n(x_n)-f(x_n)|=1\geq\varepsilon$.

In order to equate$$\lim_{n\to\infty}\int f_n(x)\text{d}x=\int\lim_{n\to\infty}f_n(x)\text{d}x=\int f(x)\text{d}x,$$it is required that $f_n$ converge to $f$ uniformly. So the OP does not suggest that $1=0$.

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  • $\begingroup$ First fixing an $x\in \mathbb{R}$ it means that $f_n$'s are uniformly continuous. Can I safely assume uniform continuity in this case? $\endgroup$
    – EEEB
    Apr 1 '19 at 17:02
  • $\begingroup$ @EEEB $f_n$ cannot be uniformly continuous since it is not even continuous. If you fix an $x\in\mathbb R$ and look at $(f_n(x))_{n=1}^\infty$, note that the preceding is a sequence of $0$s and $1$s that converges to $0$. $\endgroup$
    – wjm
    Apr 1 '19 at 17:06
  • $\begingroup$ For each $x\in \mathbb{R}$ I'm going to have a sequence $(0,0,1,0,0,\dots)$ etc (thanks for the intuition) with the position of the 1 depending on the $x$ I choose. I can see how in this case the limit is zero. My only point of objection is if I pick an $x$ extremely large ($x\to \infty)$, then the 1 can be displaced arbitrarily far away... $\endgroup$
    – EEEB
    Apr 1 '19 at 17:15
  • $\begingroup$ @EEEB That is correct; you see that to every $x$, there corresponds a sequence of $0$s and $1$s that converges to $0$. Extremely large $x$s have corresponding sequences whose finitely many $1$s are extremely far away. Nevertheless, said sequences still converge to $0$. We say therefore that the sequence of functions $(f_n)_{n=1}^\infty$ converges pointwise to the zero function. $\endgroup$
    – wjm
    Apr 1 '19 at 17:20
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    $\begingroup$ @EEEB I edited my post with more details. $\endgroup$
    – wjm
    Apr 1 '19 at 19:17

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