Show that $f_n$ converges in $L^1$ norm

Question

Suppose that $f_n: X \to C$ are a dominated sequence of measurable functions, and let $f:X \to C$ be another measurable function. Show that $f_n$ converges in $L^1$ norm to $f$ if and only if $f_n$ converges in measure to $f$.

Answer:

We know every dominated sequence of measurable functions is uniformly integrable functions.

Thus $f_n$ are uniformly integrable functions.

i.e,

for arbitrary $\epsilon>0$ there exists $\delta>0$ such that $$ \int_E |f_n| d \mu < \epsilon ,$$ whenevr $n \geq 1$ and $ E$ is measurable set with $\mu(E) < \delta$.

Similarly, $$ \int_E |f|d \mu < \epsilon. $$

In order to show $ f_n \to f$ in $L^1$ norm, we need to show that $$ \int_E f_n d \mu \to \int_E f d \mu$$.

Help me

Answer 1

Convergence in the $L^1$-norm means showing that $$\lim_{n\rightarrow\infty} \int_E |f_n - f| = 0.$$ Convergence in measure means that $$\lim_{n \rightarrow \infty} \mu(\{x \in E : |f(x) - f_n (x)| \geq \epsilon \} ) = 0.$$

Now, suppose $f_n$ converges to $f$ in the $L^1$-norm. Then, for sufficiently large n, $$\int_E |f_n - f| < \epsilon$$ which means that the average value of $|f_n - f|$ is becoming arbitrarily small. Then, since $|f_n -f| < \int_E |f_n - f|$, for large enough n, $$|f_n - f| < \epsilon_1 \rightarrow \mu(\{x \in E : |f(x) - f_n (x) | \geq \epsilon_2\}) = 0.$$

In the other direction, suppose $f_n$ converges to $f$ in measure. Then, let $A = \{x \in E : |f(x) - f_n (x)|\geq \epsilon\}$ and $B = \{x \in E : |f(x) - f_n(x)| < \epsilon\}$. Then, $$\int_E |f(x) - f_n (x)| = \int_A |f(x) - f_n(x)| + \int_B |f(x) - f_n(x)| = \int_A |f(x) - f_n(x)| + \int_B \epsilon_3 \rightarrow \int_A |f(x) - f_n(x)|.$$

To get rid of the term that's being integrated over $A$, consider using dominated convergence theorem, since you have that all your $f_n$'s are dominated.

Answer 2

For one direction, use the fact that convergence in measure implies convergence almost everywhere of a subsequence. So, if $\|f_n-f\|_1\nrightarrow 0$ , then there exists an $\epsilon > 0$ and a subsequence $(f_{n_k})$ such that $\|f_{n_k}-f\|_1>\epsilon$ for every integer $k$. Of course, $f_{n_k}\to f$ in measure as well so there is a subsequece (of this subsequence) that converges to $f$ almost everywhere. But then since $(f_{n_k})$ is dominated, we also have $\|f_{n_{k_j}}-f\|_1\to 0$ which is a contradiction because in this case, there must be an integer $j_0$ for which $\|f_{n_{k_{j_0}}}-f\|_1<\epsilon.$

Chebychev's inequality will prove the other implication:

$$\mu(|f_n-f|\ge\epsilon) \le \frac{1}{\epsilon}\int|f_n-f|\,d\mu = \frac{1}{\epsilon}\|f_n-f\|_1$$