# $f \in L^{\infty}(\mathbb{R}), ||f_h - f||_{\infty} \rightarrow 0 \implies f = g$, uniformly continuous, a.e.

The following question has been asked before, and I'm reasking to get feedback on my attempt.

Question: Suppose $$f \in L^{\infty}(\mathbb{R})$$, $$f_{h}(x) = f(x+h)$$, and $$\lim_{h \rightarrow 0}||f_{h} - f||_{\infty} = 0$$ Prove that there exists a uniformly continuous function $$g$$ on $$\mathbb{R}$$ such that $$f = g$$ a.e

Attempt: Let us first think about the case where $$f$$ is bounded everywhere and $$f_h \rightarrow f$$ uniformly on $$\mathbb{R}$$. Then, we have that $$\sup_{x \in \mathbb{R}} |f(x+h)-f(h)| \rightarrow 0$$, which is the definition of $$f$$ being uniformly continuous, so $$g := f$$ works. (Note here that we don't use the bounded everywhere assumption)

Now, to address the general case, I hope to construct a set $$A$$ such that $$m(A) = 0$$, and $$f_h \rightarrow f$$ uniformly on $$A^c$$. Then, by the above simple case, $$f$$ is uniformly continuous on $$A^c$$, and we are done.

By assumption that $$||f_h - f||_{\infty} \rightarrow 0$$, i.e., $$\lim_{h \rightarrow 0} \inf \{M: m(|f_n - f| > M) < \epsilon \} \rightarrow 0$$, we have that $$\forall \epsilon, \eta > 0, \exists \delta > 0$$, s.t. $$|h|< \delta_k \implies m(|f_h - f| > \eta) < \epsilon .$$ Choosing $$\epsilon_k := 2^{-k}, \eta_k := \frac{1}{k}$$ and satisfactory $$\delta_k$$ such that $$|h| \leq \delta_k \implies m(|f_h - f| > \eta_k) < \epsilon_k$$, and letting $$A_k := \{|f_{\delta_k}- f| > \eta_k \}$$, we have that $$m(\limsup A_k) = 0$$ by Borel-Cantelli (I).

Hence, if $$A := \limsup A_k$$, then $$x \in A^c \implies x \in A_k \text{ almost always} \implies x \in A_n$$ whenever $$n \geq N(x) \in \mathbb{N}$$, i.e., $$|f_{\delta_n}(x)- f(x)| \leq \eta_n$$.

Question

1. I'm not sure how to extend the above to show that $$|f_h(x) - f(x)| < \epsilon$$ for $$h$$ sufficently small (i.e., $$f_h \rightarrow f$$ pointwise on A^c).

2. I'm also not sure how to extend the above to get that $$f_h \rightarrow f$$ uniformly on $$A^c$$.

3. Lastly, even if the above works, is it true that (via the simple case) we have $$f$$ uniformly continuous on $$A^c$$, and we are done? If so, where did we use the fact that $$f \in L^{\infty}$$?

• Hints: $\lim \sup_n f_{1/n}$ is (finite and) uniformly continuous on dense set, so it extends to a uniformly continuous function $g$ on $\mathbb R$. Commented Aug 1 at 7:50
• I haven't been able to show that $f_{\frac{1}{n}}$ converges uniformly to its limit though: any suggestions on that @geetha290krm? Commented Aug 1 at 11:51