Prove $f_n\to f$ on $[a,b]\implies \int_a^b|f_n-f|\to 0$

Question

Suppose $f,f_n$ are measurable and uniformly bounded on $[a,b]$. Prove $f_n\to f$ on $[a,b]\implies \int_a^b|f_n-f|\to 0$

Attempt:

We note that since $f$ and $f_n$ are bounded and are measurable, they are both Lebesgue integrable. Suppose $f_n\to f$ on $[a,b]$. We can write this as $|f_n-f|<\epsilon$ on $[a,b]$ whenever $n\geq N$. Then for $\forall n\geq N$, we have that $\int_a^b |f_n-f|<\int_a^b \epsilon=\epsilon (b-a)$ from the monotonicity of the integral.

Edit 1:

(Attempt 2)

Since $f_n\to f$on $[a,b]$, a measurable set with finite measure, the hypotheses of Egorov's Theorem are met. So, by Egorov's Theorem we know $\exists A\subset [a,b]: m(A)<\epsilon$ and $f_n\to f$ uniformly on $[a,b]\setminus A$.

So we know that $\forall n\geq N$, we have $|f_n-f|<\epsilon$ on $[a,b]\setminus A$. Taking the integral of both sides, we get (due to monotonicity of the integral) that $\int_a^b|f_n-f|<\int_a^b\epsilon=(b-a)\epsilon$ on $[a,b]\setminus A$, but since $m(A)<\epsilon$ we have that $\int_a^b |f_n-f|$ on $[a,b]\setminus A$ is equal to $\int_a^b|f_n-f|$ on $[a,b]$.

(Attempt 3)

Proof.

Suppose $(f_n)$ is a measurable sequence of functions such that $f_n\to f$. Furthermore, since $f_n$ is uniformly bounded on $[a,b]$, we have $|f_n|\leq M\in\mathbb{R}$. The hypotheses of Egorov's Theorem are met, so $\exists A\text{ measurable }\subset [a,b]: m(A)<\epsilon$ and $f_n\to f$ uniformly on $[a,b]\setminus A$. Since we have uniform convergence, it follows that $\forall\epsilon>0\exists N\in\mathbb{N}:\forall n\geq N\implies |f_n-f|<\epsilon$ on $A$.

Then we get

\begin{align*} \left|\int_{[a,b]} (f_n-f)\right|&\leq \int_{[a,b]}|f_n-f|\\ &=\int_{[a,b]\setminus A}|f_n-f|+\int_{ A}|f_n-f|\\ &<\int_{[a,b]\setminus A}2M+\int_A \epsilon\\ &<\int 2M\cdot \chi_{[a,b]\setminus A}+\int \epsilon \cdot \chi_{A}\\ \end{align*}

But I'm having some difficulty simplifying this to show the desired result (if this is the right way).

Thanks!

Answer 1

Solution 1: (This makes attempt 1 rigorous.) For fixed $\epsilon>0$, we have

$$\int_a^b |f_n-f| \, dx = \int_{[a,b] \cap \{|f_n-f|>\epsilon\}} |f_n-f| \, dx + \int_{[a,b] \cap \{|f_n-f| \leq \epsilon\}} |f_n-f| \, dx=: I_1+I_2.$$

1. Show that $I_2 \leq \epsilon (b-a)$.
2. Using the uniform boundedness, show that there exists a constant $C>0$ such that $$I_1 \leq C \lambda(\{|f_n-f|>\epsilon\} \cap [a,b]).$$ Here $\lambda$ denotes the Lebesgue measure.
3. Conclude from the fact that $f_n \to f$ that $$I_1 \leq C \epsilon$$ for $n \geq N$ sufficiently large.
4. Conclude.

Solution 2:

1. By uniform boundedness there exists a constant $M>0$ such that $$g_n := -|f_n-f|+M$$ is a non-negative sequence.
2. Apply Fatou's lemma to $(g_n)_{n \in \mathbb{N}}$ to obtain $$\limsup_{n \to \infty} \int_a^b |f_n-f| \, dx \leq 0.$$
3. Conclude.

Answer 2

Suppose $(f_n)$ is a measurable sequence of functions such that $f_n\to f$. Furthermore, since $f_n$ is uniformly bounded on $[a,b]$, we have $|f_n|\leq M\in\mathbb{R}$. The hypotheses of Egorov's Theorem are met, so $\exists A\text{ measurable }\subset [a,b]: m([a,b]\setminus A)<\epsilon$ and $f_n\to f$ uniformly on $[a,b]\setminus A$. Since we have uniform convergence, it follows that $\forall\epsilon>0\exists N\in\mathbb{N}:\forall n\geq N\implies |f_n-f|<\epsilon$ on $A$.

Then we get

\begin{align*} \left|\int_{[a,b]} (f_n-f)\right|&\leq \int_{[a,b]}|f_n-f|\\ &=\int_{[a,b]\setminus A}|f_n-f|+\int_{ A}|f_n-f|\\ &<\int_{[a,b]\setminus A}2M+\int_A \epsilon\\ &<2M\cdot m([a,b]\setminus A)+\epsilon\cdot m([a,b])\\ &<(2M+m([a,b]))\epsilon \end{align*}

But since $m([a,b])=b-a<\infty$, we know that $\forall\epsilon>0\exists N\in\mathbb{N}:\forall n\geq N\implies |\int _{[a,b]}f_n-f|<c\epsilon$ for $c\in\mathbb{R}$. So finally get $\int_{[a,b]}f_n\to \int_{[a,b]} f$ which implies our desired result: $\int^a_b|f_n-f|\to 0$.