$L^1$ convergence and absolute continuity

Question

I'm currently studying for a real analysis qual and came across this question:

Let $[a, b]$ be a bounded interval of $\mathbb R$ and let $m$ be Lebesgue measure. Let $M$ be a positive real number, and let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on $[a, b]$ such that $\int_a^b |f_n| dm \leq M$ for every $n$. Assume that $f_n(x) \to f(x)$ as $n \to \infty$ for $m$-almost every $x$.

a) State Fatou's Lemma.

b) Show that $\int_a^b |f| dm \leq M$.

c) Suppose that $\|f_n - f\|_1 \to 0$. Prove for every $\epsilon > 0$ there exists $\delta > 0$ such that if $A \subset [a, b]$ is $m$-measurable and $m(A) \leq \delta$, then $\int_A |f_n| dm \leq \epsilon$ for all $n$.

I'm okay with parts (a) and (b), but part (c) is giving me trouble. My thought is that I can use absolute continuity: $f \in L^1(\mu)$ implies $\exists \delta > 0$ such that $|\int_E f d\mu| < \epsilon$ for $\mu(E) < \delta$ (ref: Folland 3.6). But that gives rise to a bunch of questions:

• Does that say anything about $\int_E |f| d\mu$? I unsuccessfully tried to apply the triangle inequality in a few different ways.

• I don't know if $f \in L^1(\mathbb R, m)$ or $f_n \in L^1(\mathbb R, m)$. Can I instead say something like $f_n \in L^1([a, b], m)$? Does that even make sense? Because then I could conclude $\left|\int_A |f_n|dm\right| \leq \epsilon$

• If that works, then I haven't made use of the hypothesis of $L^1$ convergence, but that seems important.

• For that matter, I'm not totally sure if / how this follows from parts (a) and (b).

Basically, I have a lot of vague thoughts floating around, so any advice for a specific direction to go in would be super helpful!

Answer 1

For you first bullet, you actually have $∫_E |f| \mathrm{d} < $, not just $|∫_E f \mathrm{d}| < $. You can see that from your previous result by considering the function $|f∣$ instead of $f$ (as said in the comments).

For the second bullet, yes, it makes sense.

If you apply the result for each $n∈ℕ$, you get a $_n$ (depending on $n$) such that $∫_A |f_n| < $ if $m(A) < _n$. But these $_n$ can have an infimum of $0$, so you cannot take this infimum to get a $$ working for all $n$. But for every finite family of natural number, it works. So we have proved the result for all interval $\{0,1,…,N\}$ instead of all $ℕ$ (we now have to treat what happens at infinity).

We have to find an $N∈ℕ$ and a $>0$ such that $∫_A |f_n| < $ if $m(A)<$ and $n≥N$. You have to use that $‖f_n-f‖_1 → 0$ to make $f_n$ near enough to $f$. Then, apply the result (Folland 3.6) to $f$ and try to “transfer” it to each measurable function near enough to $f$ (but you will have a bigger $$ depending on the distance at $f$ you allow). I write what I have in mind below if you want.

Suppose $‖g-f‖ < $. It means that $∫_a^b |g-f| < $. If we restrict to a subset $A⊆[a,b]$, the integral can only decrease since the integrand is positive: $∫_A |g-f| < $. We get $∫_A |g| < + ∫_A |f|$. So by forcing $n$ to be big enough, we can control $∫_A |f_n|$ by $∫_A |f|$. But by the result (Folland 3.6) we can make $∫_A |f|$ small by making $A$ small.
More precisely, let $N$ be such that $‖f_n-f‖</2$ for all $n≥N$. Let $_∞$ be such that $∫_A |f| < /2$ if $m(A) < _∞$. Then for all $n≥N$ and all $A$ with $m(A) < $, we have $∫_A |f_n| < /2 + /2 = $.
After that, for each $i<N$, take $_i$ such that $∫_A |f_i| < $ if $m(A) < _i$. The final $$ is $\min(_0,_1,…,_{N-1},_∞)$.

I don't see the relation with parts (a) and (b) either.

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If you are interested, we can formulate that more abstractly using the notion of compactness ($ℕ∪\{∞\}$ behave like a finite set). The two say the same thing, we can make the proofs correspond.

Let $M$ be the set of measurable regions. For $A,B∈M$, the distance between $A$ and $B$ is $m(AB)$, where $$ is the symmetric difference. (It is like a metric space.) Then we have a continuous function $M×L^1 → ℝ_{≥0}$ sending $(A,f)$ on $∫_A |f|$ (it is $1$-lipschitz in $f$ and continuous in $A$, this is the important thing to prove). The sequence $(f_n)_n$ converging to $f$ is a continuous function from $\overline ℕ$ to $L^1$. This gives a continuous function $M×\overline{ℕ} → ℝ_{≥0}$. The subspace $\{∅\}×\overline{ℕ}$ is sent on $0$. So by the tube lemma, for all $>0$, there is $>0$ such that $\{A∈M\,|\,m(A)<\}×\overline{ℕ}$ is sent in $\left]-,\right[$.

Answer 2

Part b) tells you $f\in L^1[a,b].$ You already knew all $f_n$ are in $L^1[a,b].$ (Not sure why you ask if being in $L^1[a,b]$ even makes sense. Of course it does!)

Hint for c): Let $\epsilon >0.$ By the absolute continuity property you mentioned, we can find a $\delta>0$ that works for $f,$ and for each $n,$ a $\delta_n>0$ that works for $f_n.$ So how about setting $\delta = \inf\, \{\delta_n: n = 0,1,2,\dots \}?$ That's no good because that infimum might be $0$ for all we know. But wait: You also you know $\int_a^b |f_n-f| <\epsilon$ for $n>$ some $N.$ That's huge!