$\int_{\Omega} |f_n-f||f_n| \, d \mu \to 0$ if $f_n \in L^1(\Omega)$, $f_n \to f$.

Question

Suppose $f_n \to f$ in $L^1(\Omega)$ where $\mu(\Omega)=1$. Suppose $$\int_{\Omega} |f_n| \, d\mu \leq M$$ for all $n$.

Is there a way to show that the integral

$$\int_{\Omega} |f_n-f||f_n| \, d \mu \to 0$$ as well?

I was thinking of defining a set $A_N$ where given $N$, $A(N) := \{ x \in \Omega : f_n (x) \geq N \, \text{for any} \,n \in \mathbb N \}$ and showing that $$\int_{\Omega} |f_n-f||f_n| \, d \mu \leq N\int_{\Omega \setminus A(N)} |f_n-f| \, d \mu + \int_{ A(N)} |f_n-f||f_n| \, d \mu$$ goes to $0$, but I'm not sure if I can show this. The thought is that $\mu(A(N))$ must get arbitrarily small as $N$ increases.

Ideas? Hints? Could I use the fact that if $h \in L^1$, then there exists $\delta$ so that if $\mu(E)<\delta$ then $\|h\|_1 <\epsilon$.

Answer 1

In general, we cannot expect that

$$\int_{\Omega} |f_n-f| \cdot |f_n| \, d\mu \to 0;$$

roughly, the reason is that $L^1$-convergence does not imply $L^2$-convergence.

(Counter)Example Consider $((0,1),\mathcal{B}((0,1)))$ endowed with the Lebesgue measure restricted to $(0,1)$ and

$$f_n(x) := n^{3/4} 1_{[0,1/n]}(x).$$

Then

$$\int |f_n(x)| \, dx = n^{-1/4} \xrightarrow[]{n \to \infty} 0$$

i.e. $f_n$ converges in $L^1$ to $f=0$. This also implies that $\sup_{n \in \mathbb{N}} \|f_n\|_{L^1}<\infty$. On the other hand,

$$\int |f_n(x)-f(x)| \cdot |f_n(x)| \, dx = \int |f_n(x)|^2 \, dx = n^{1/2}$$

does not converge to $0$ as $n \to \infty$.