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I was solving an exercise problem in Rudin's Real and Complex Analysis when I came across this statement.

Construct an example in which Vitali's theorem applies although the hypotheses of Lebesgue's theorem [DCT] do not hold.

I am really wondering what such an example would be. UI requires $\mathcal{L}^1$ and boundedness, while DCT requires boundedness by another function. I am unable to segregate the 2 different conditions. Could someone point me to an example that behaves as stated by Rudin?

Vitali convergence theorem: $(X,\mathcal{F},\mu)$ is a positive measure space. If a) $\mu(X)<\infty$ b) $\{f_n\}$ is uniformly integrable c) $f_n(x)\to f(x)$ a.e. as $n \to \infty$ and d) $|f(x)|<\infty$ a.e. then 1) $f\in \mathcal{L}^1(\mu)$ 2) $\lim_{n\to \infty} \int_{X}|f_n-f|d\mu=0$.

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up vote 9 down vote accepted

One idea that proves fruitful is to "chop up" a function that is not in $L_1[0,1]$ into pieces $f\cdot \chi_{A_k}$ where the measure of the $A_k$ are small enough to insure uniform integrability of the sequence and in such a way that the sequence converges pointwise:

Let $n$ be a positive integer and set $A_n=[2^{-(n+1)}, 2^{-n}]$. Partition $A_n$ into $n$ measurable sets $A_n^1$, $A_n^2$, $\ldots\,$, $A_n^n$ each of measure $1/(n\cdot2^{n+1})$. For $1\le i\le n$, define the function $f_n^i$ via $f_n^i(x) ={1\over x}\chi_{A_n^i}$.

Now consider the sequence $S=(f_1^1, f_2^1, f_2^2, f_3^1, f_3^2,f_3^3,f_4^1\ldots)$.

To see that this sequence is uniformly integrable:

Let $n$ be a positive integer. Set $\delta={1\over n2^{n+1}}$. Suppose $\mu(E)<\delta$. Then for any $m\ge n$ and $1\le i\le m$, we have $$ \int_E |f_m^i| \le \int_{A_m^i}|f_m^i| \le 2^{m+1}\cdot {1\over m2^{m+1}}= {1\over m}<{1\over n}. $$ While for $k<n$ and $1\le i\le k$, we have $$ \int_E |f_k^i| \le \delta\cdot 2^{k+1}={2^{k+1}\over n2^{n+1}}<{1\over n} . $$ Thus, $S$ is uniformly integrable.

By construction of the $A_n$, it is easily seen that $S$ converges pointwise to 0.

So, $S$ satisfies the hypotheses of the Vitali Convergence Theorem. But $S$ is not dominated by any $L_1$ function, since any dominating function would have to dominate $f(x)={1\over x}\notin L_1$.

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Much simpler would be to take $f_n(x)=(1/x)\chi_{(1/n,1/(n+1)]}$. – David Mitra Jul 2 '14 at 20:44

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