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I'm always a bit confuse with this convergence in distribution sense. For example, take $f_n(x)=n\boldsymbol 1_{[0,1/n]}$. We have that $f_n(x)\to 0$ a.e., so why do we say that $f_n\to \delta _0$ in distribution ? I don't really understand the subtlety (and by the way it looks a bit the same except that $\delta _0(0)=1$ whereas $\lim_{n\to \infty }f_n(0)=\infty $. But, I saw on wikipedia that $\delta _0(0)=\infty $ by convention, so maybe, at the end it works...

Anyway, I don't understand this convergence in distribution sense, at least, in this example, it looks the same...

user657324
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1 Answers1

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I think my answer to your question in the previous link should help. $f_n(x)\to f(x)$ doesn't implies that $\int f_n\to \int f$. Notice that Fatou lemma say that $$\int\liminf_{n\to \infty }|f_n|\leq \liminf_{n\to \infty } \int |f_n|,$$ but no more, and your sequence $(f_n)$ is an example of sequence such that the inequality is strict.

Moreover, $\delta _0$ is not a function since there are no functions such that $f=0$ almost everywhere but $\displaystyle\int f=1$.

Some result :

  • If $f_n(x)\to f(x)$ almost everywhere and $f_n,f$ are locally $L^1$, then $(f_n)$ converge to a distribution (but you don't necessary have $f_n\to f$ in distribution)

  • If $f_n(x)\to f(x)$ almost everywhere and $\int |f_n|\to \int |f|$, then $f_n\to f$ in $L^1$.

  • If $f_n\to f$ weakly, then $f_n\to f$ in distribution sense.

user659895
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