Let $X = \mathbb{R}$ and $M$ denote the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$, while $m$ denotes the outer Lebesgue measure restricted to $M$.

Consider $f_n = \chi_{(1,2]}$ (characteristic function of $(1,2])$ when $n$ is even and $f_n = \chi _{[0,1]}$ when $n$ is odd. Since both $(1,2]$ and $[0,1]$ are Lebesgue measurable, we have that for each $n$, $f_n$ is a measurable function.

Now when $n$ is odd, $\int_X f_n dm = m(1,2] = 1$ and $m[0,1] = 1$ for even $n$. Hence for each $n$ the integral is $1$.

Hence, $\lim \inf \int_X f_n dm = 1$. However clearly, $\lim\inf f_n = 0$. So $\int_X \lim \inf f_n dm = 0$. Hence the inequality in Fatou's lemma can be strict.

Is this a correct argument, or have I made a mistake somewhere?

  • $\begingroup$ It is correct (but there is a typo in the definition of $f_n$). $\endgroup$ – Rigel Jun 22 '17 at 15:08
  • $\begingroup$ The only mistake I see is a typo: When $n$ is even, $f_n$ was intended to be $\chi_{[0,1]}$. $\endgroup$ – Andreas Blass Jun 22 '17 at 15:09
  • $\begingroup$ Oh yes, I see. Thanks for the feedback :) $\endgroup$ – fosho Jun 22 '17 at 15:10

For a sequence of functions $\left(f_n\right)_{n\geqslant 1} $, let $$A:=\int\liminf_{n\to\infty}f_n\mathrm d\lambda \mbox{ and }B:=\liminf_{n\to\infty}\int f_n\mathrm d\lambda .$$ Fatou's lemma states that we always have $A\leqslant B$, and the (correct) proof in the opening post show that the inequality may be strict. We may also have $A=0$ and $B= +\infty$, for example if $f_n=\mathbf 1_{\left[2^n,2^{n+1}\right)}$.


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