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

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