Finding dominating function

Question

Let $f_n:(0,1)\rightarrow\mathbb{R}$ be $n$ for $x\in(0,1/n)$ and be $0$ otherwise. Then why this $f_n$ have no dominating function which is lebesgue integrable?

I want that dominating function condition cannot be cancelled from dominated convergence theorem

Answer 1

So we have

$$\forall n, \int_{\mathbb{R}} f_n =1$$

and in the meaning of distribution we get Dirac distribution

$$ f_n \to \delta_0 $$

which is not Lesbegues integrable.

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Note that if we get domination we would get $\delta_0$ integrable which is contradictory.

Answer 2

Suppose $g(x) \geq f_n(x)$ for all $x$ and all $n$. Then $g(x) \geq n$ for $\frac 1 {n+1} <x<\frac 1 n$. Hence $\int_0^{1} g(x) dx \geq \sum \int_{1/n+1}^{1/n} g(x)dx \geq \sum n(\frac 1 n-\frac 1 {n+1})=\sum \frac 1 {n+1}=\infty$.