$f_j \rightarrow f$ in $L_1$ then $\mathcal{F}(f_j) \rightarrow \mathcal{F}(f)$ in measure

Question

Assume that $f_j \in L_1(\mathbb{R}^n)$ for all j. Does it follow that if $f_j \rightarrow f$ in $L_1$ then $\mathcal{F}(f_j) \rightarrow \mathcal{F}(f)$ in measure? I don't think this is true but I can't think of a counterexample.

Answer 1

Note that the Fourier transform $\mathcal{F}g$ of a function $g$ satisfies $$\|\mathcal{F}g\|_{\infty} := \sup_{\xi \in \mathbb{R}^d} |\mathcal{F}g(\xi)| \leq \|g\|_{L^1}.$$ (Possibly with an additional multiplicative constant on the right-hand side, depending on your definition of the Fourier transform.) Consequently, $$\|\mathcal{F}f_j - \mathcal{F}f\|_{\infty} = \|\mathcal{F}(f_j-f)\|_{\infty} \leq \|f-f_j\|_{L^1} \xrightarrow[]{j \to \infty} 0$$ for any sequence $f_j \to f$ in $L^1$. This shows that $\mathcal{F}f_j$ converges uniformly to $\mathcal{F}f$; hence, in particular, in measure.

Remark: If $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\mu)$ is a measure space and $\mu$ a measure which is not finite, then pointwise convergence does, in general, not imply convergence in measure. However, uniform convergence implies convergence in measure. Indeed: Suppose that $g_j$ converges uniformly to $g$. For fixed $\epsilon>0$ we can choose $N \in \mathbb{N}$ sufficiently large such that $|g_j(x)-g(x)| \leq \epsilon$ for all $j \geq N$, $x \in \mathbb{R}^d$. This implies $$\mu(|g_j-g|>\epsilon) = \mu(\emptyset)=0 \qquad \text{for all $j \geq N$};$$ hence, $g_j \to g$ in measure.