# Showing that $f_n \to f$ in $L^1{[0,1]}$

Suppose

(a) $f_n \to f$ a.e. on $[0,1]$ and
(b) $\|f_n\|_{L^\infty_{(0,1)}} \leq M \lt \infty.$

I want to show that $f$ is Lebesgue integrable and $f_n \to f$ in $L^1{[0,1]}$.

Also, I want an example where the condition (b) is not satisfied and the conclusion fails.

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That $f_n\to f$ in $L^1$ follows from Lebesgue's dominated convergence theorem.

For the counterexample, let $f_n(x)=n\ \chi_{[0,1/n]} (x)$, where: $$\chi_{[0,1/n]} (x):=\begin{cases} 1 &\text{, if } 0\leq x\leq 1/n \\ 0 &\text{, otherwise.}\end{cases}$$ Then $f_n\to 0=:f$ a.e. in $[0,1]$ but $\lVert f_n\rVert_\infty\to \infty$ and: $$\int_0^1 |f_n| = 1 \qquad \text{and} \qquad \int_0^1 |f| =0\; ,$$ therefore $f_n\not\to f$ in $L^1$.

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Thanks very much for the counterexample. – Kuku Mar 13 '12 at 23:20
Could you help me prove the first part. Thanks. – Kuku Apr 7 '12 at 12:49

Part (a) follows directly from the Dominated Convergence Theorem since the constant function $M$ is in $L^1$. For the second part define $f_n(x)$ to be zero on the interval $[1/n,1]$ and make the function large enough so that $\int_0^{1/n} f_n(x)dx=1$.

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Thanks for your response. – Kuku Mar 13 '12 at 23:19
How do I interpret condition (b)? – Kuku Mar 13 '12 at 23:52

Part (a) follows directly from Lebesgue's Dominated Convergence Theorem, where the dominating function is the constant function equal to $M$.

http://en.wikipedia.org/wiki/Dominated_convergence_theorem

For part (b), consider a function that is 0 everywhere outside $[1/n,1]$ but which "spikes" at $\frac{1}{2n}$ to a height of $2n$. (ie, let $f(x)=4n^2x$ when $x\leq 1/2n$ and $f(x)=4n^2(1/n-x)$ when $1/2n\leq x\leq 1/n$.) Then $\int f_n=1$ for all $n$ but $\int f=0$.

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Thanks for your response. What if I want to use Fatou's Lemma for the first part. – Kuku Mar 13 '12 at 23:19
Then I would mimic the proof of dominated convergence useing Fatou's lemma. – Brett Frankel Mar 13 '12 at 23:34