One can easily apply the dominated convergence theorem (DCT) to see that there is no integrable dominating function for each of the following functions

  • $f_n=1_{[n,n+1]}$,
  • $h_n=\frac{1}{n}1_{[0,n]}$,
  • $g_n=n1_{[1/n,2/n]}$.

For "integrable dominating function" I mean the dominating function in the assumption of the DCT. I would like to see why in each of these examples, no dominating function exists without appealing to the DCT.

If $G\geq f_n$ for all $n$, then $G\geq 1$ a.e..

If $G\geq h_n$ for all $n$, then $$ \int G\geq \sum_{n=1}^\infty\frac{1}{n}. $$

Could anyone help me with the third one?


For each $n$, $$ g_n(x):= nI_{\left[\frac1n, \frac2n\right]}(x)\ge \frac1x I_{\left[\frac1n,\frac2n\right]}(x)\ge \frac1xI_{\left(\frac2{n+1},\frac2n\right]}(x). $$ Therefore if $G(x)\ge g_n(x)$ for all $n$, then $G(x)\ge\frac1x$ on the interval $(0,2]$.


If $\forall n,\ G\ge g_n$, then $\forall n,\ G\ge g_{2^n}$. Since $[2^{-n},2^{-n+1}]\cap [2^{-m},2^{-m+1}]$ is negligible whenever $m\ne n$, you get that $$\sup_{n\in\mathbb N} g_{2^n}=\sum_{k=1}^\infty g_{2^k}$$ hence

$$G\ge \sum_{k=1}^\infty g_{2^k}$$

But then $$\int G\,dx\ge \sum_{k=1}^\infty 2^k\cdot 2^{-k}=+\infty$$


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