I was recently told that the "Big Three" convergence theorems for the Lebesgue integral (Fatou, dominated, and monotone) are equivalent. I'm trying to show this directly by writing six proofs instead of the usual three. I'm currently stuck on the "dominated convergence theorem implies Fatou" part.

Suppose that the dominated convergence theorem holds, and let $\{f_n\}$ be a sequence of nonnegative, measurable functions. How can we show that $$\int \liminf f_n \leq \liminf \int f_n$$ using the dominated convergence theorem?

My first attempt was this: Set $g_n = \inf_{k \geq n} f_k$, so that $g = \lim g_n = \liminf f_n$. This gives us $$g_1 \leq g_2 \leq \cdots \leq g,$$ so if $g$ is integrable, we can apply the DCT to get the result. However, if $g$ is not integrable, then I see no way forward.

This is related to this unanswered question, but the comments there only point out the problem with this proof.

  • $\begingroup$ Per definiton of the limes inferior we have that $\text{inf}_{m:m \geq n}f_m \uparrow \lim\inf_{n\rightarrow \infty} f_n$. Try using DCT. $\endgroup$ Feb 15 '19 at 22:27
  • $\begingroup$ @babemcnuggets Try using the DCT on what sequence? On $\inf_{m \geq n} f_m$? What if $\liminf f_n$ isn't integrable? $\endgroup$
    – Robert D-B
    Feb 15 '19 at 22:55
  • $\begingroup$ You would have that $\int \lim \inf_{n\rightarrow\infty} f_n\text{d}\mu\space =\space \lim_{n\rightarrow \infty} \int \inf_{m:m\geq n}f_m \text{d}\mu$ using monotone Convergence. Because of the equality it is integrable because the right side is. $\endgroup$ Feb 15 '19 at 23:28
  • $\begingroup$ @rwbogl I seriously doubt if a single application of DCT gives Fatou's Lemma immediately. $\endgroup$ Feb 16 '19 at 12:15

$\int g_n \geq \int_{g_n >(1-\epsilon)g,g>0} g_n\geq (1-\epsilon)\int_{g_n >(1-\epsilon)g,g>0} g \to \infty$ if $\int g=\infty$ (because the sets $\{g_n >(1-\epsilon)g,g>0\}$ increase to the set $\{g>0\}$. This proves $\int g_n \to \int g$ whenever $\int g=\infty$. When $\int g <\infty$ we can apply DCT.

Thus $\int \lim \inf f_n=\int g=\lim \int g_n \leq \lim \inf \int f_n$ because $g_n \leq f_n$ for all $n$.

Note: the fact that $\int_{A_n} g \to \int g$ for any non-negative measurable function $g$ when $A_n$ increase to the whole space can be proved easily by the definition of integral and this does not require any of the basic limit theorems of measure theory.

  • $\begingroup$ $\epsilon \in (0,1)$ is fixed in this proof: We may take $\epsilon =\frac 1 2$. $\endgroup$ Feb 16 '19 at 12:09

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