We know that for $p\ge1$ , given a positive measure $\mu$ on a set $X$ if $f_n\to f$ almost everywhere, $f,f_n\in L^p(\mu)$ for all $n$, and $||f_n||_p\to ||f||_p$ as $n\to \infty$ then $||f-f_n||_p\to 0$ (See here)

Will this remain true for $0<p<1$?

The inequality we get here is $|f-f_n|^p<|f|^p+|f_n|^p$. I found this problem in big Rudin, where he suggests: By using Egorov's theorem write $X=A \cup B$ such that $\int_A |f|^p \le \varepsilon$, and $\mu(B)\le \infty$ and $f_n\to f$ uniformly on $B$ and then applying Fatou's lemma to $\int_B|f_n|^p$. This doesn't seem like a direct application of Egorov's since the space is not given to have finite measure.


2 Answers 2


It's an indirect application of Egorov's theorem. You first construct a situation where you can apply Egorov's theorem. Let $C_n = \{x : \lvert f(x)\rvert^p \geqslant 2^{-n}\}$. Then since $\lvert f\rvert^p$ is integrable, all $C_n$ have finite $\mu$-measure. For large enough $n$, you have

$$\int_{X\setminus C_n} \lvert f\rvert^p\,d\mu < \frac{\varepsilon}{2}.$$

There is a $\delta > 0$ such that for every measurable $E\subset X$ with $\mu(E) \leqslant \delta$ you have

$$\int_{E} \lvert f\rvert^p\,d\mu < \frac{\varepsilon}{2}.$$

Now let $B$ be a subset of $C_n$ such that $\mu(C_n \setminus B) \leqslant \delta$ and $\lvert f_n\rvert^p$ converges uniformly to $\lvert f\rvert^p$ on $B$ (by Egorov's theorem), and for $A = X\setminus B = (X\setminus C_n) \cup (C_n\setminus B)$ we have $$\int_A \lvert f\rvert^p\,d\mu < \varepsilon$$ by construction. Then use Fatou's lemma as indicated.

  • $\begingroup$ I don't see why $\int_{X\setminus C_n} \lvert f\rvert^p\,d\mu < \frac{\varepsilon}{2}$ , I see that $\int_{X\setminus C_n} \lvert f\rvert^p\,d\mu \le \int_{X\setminus C_n} \lvert 2^{-n}\rvert^p\,d\mu$, but how do we get $\varepsilon$, and how do we deduce the existence of $\delta$? $\endgroup$
    – user10444
    Nov 1, 2014 at 23:10
  • $\begingroup$ The $C_n$ form an increasing sequence of measurable sets with $\bigcup C_n = \{ x : f(x) \neq 0\}$. Hence by the dominated convergence theorem, $$\int_{C_n} \lvert f\rvert^p\,d\mu = \int_X \lvert f\rvert^p \cdot \chi_{C_n}\,d\mu \to \int_X \lvert f\rvert^p\,d\mu,$$ and therefore, for all large enough $n$, the integral over $X\setminus C_n$ is smaller than $\varepsilon/2$. $\endgroup$ Nov 1, 2014 at 23:18
  • $\begingroup$ $\varepsilon > 0$ is arbitrarily given. The construction shows that for all $\varepsilon > 0$, you can find an $n_0$ such that for all $n \geqslant n_0$ you have $\int \lvert f_n -f \rvert^p\,d\mu < 2\varepsilon$. The existence of $\delta$, given $\varepsilon$, is exercise 12 in chapter 1. $\endgroup$ Nov 1, 2014 at 23:27
  • $\begingroup$ That is a pretty beautiful way to use that exercise! $\endgroup$
    – user10444
    Nov 1, 2014 at 23:31

Seems it's still a direct application of Fatou's lemma:

Since $|f|^p + |f_n|^p - |f-f_n|^p \geq 0$, we can apply Fatou's lemma:

$$\int \liminf \left(|f|^p + |f_n|^p - |f-f_n|^p\right) \leq \liminf \int \left(|f|^p + |f_n|^p - |f-f_n|^p\right) $$


$$\int 2|f|^p \leq \int 2|f|^p - \limsup\int |f-f_n|^p $$

So $\limsup\int |f-f_n|^p \leq 0$, i.e. $\lim\int |f-f_n|^p =0$


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