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I had this problem in my midterm exam the last week and still I can't handle it:

It is true that if $f \in L^{1}(\mathbb{R})$, (that is, $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue measurable, $\int f d \lambda < \infty$, where $\lambda$ is the Lebesgue measure), and $a_{n} \subset \mathbb{R}$ is a sequence such that $a_n \rightarrow 0$, then $$ \lim_{n \rightarrow \infty} \int \mid f(x +a_n) -f(x) \mid d\lambda = 0 ?$$ This is trivially true if $f$ is continuous $\lambda$- almost everywhere. For the general case I tried to use the Dominated Convergence Theorem over $f_n:= f(x +a_n) -f(x)$, but It seems false that $f_n \rightarrow 0$ $ \lambda$-a.e. even if $f = \chi_E$ is a characteristic function where E $\in \mathcal{L}$ is a Lebesgue measurable set. Please someone can give me some pointers on this problem? Thanks in advance!

My background: We are following Folland's Real Analysis Book, the last thing we saw was modes of convergence.

Clayton
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user604739
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    This https://math.stackexchange.com/questions/59510/measure-theory-convergence-theorems-for-non-discrete-indexing-parameter ? – Martin R Mar 16 '19 at 17:04
  • You can prove it for continuous functions and then use the density of $C_c(\mathbb{R}^n)$ in $L^1$ for the general case. – GReyes Mar 16 '19 at 17:05
  • But if I want to use the density of continuous functions... I know that there exist some $g_n$ continuous arbitrary close to $f(x+a_n)$ and some $g$ arbitrary close to $f(x)$, how do I know that $g_n$ and $g$ are also very close one of each other? – user604739 Mar 16 '19 at 17:14
  • Don't use $g_n;$ find one continuous $g$ such that $|g-f|_1<\epsilon.$ – zhw. Mar 16 '19 at 17:23
  • So if $g$ is continuous and it is very close to $f$ this implies that, for sufficiently large n, $g(x+an)$ is close to $f(x+an)$? – user604739 Mar 16 '19 at 17:40
  • @ GReyes The fact that $C_c(\mathbb{R})$ is dense in $L^1$ requires a lot of work to prove. I do not know if this fact can be taken as granted. – Danny Pak-Keung Chan Mar 16 '19 at 19:46
  • @JuanuPE What facts can be taken as granted? What facts cannot? – Danny Pak-Keung Chan Mar 16 '19 at 19:49

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Both the fact that $C_{c}(\mathbb{R})$ is dense in $L^{1}(\mathbb{R})$ and the proposition you are trying to prove rely on the regularity of Lebesgue measure. To avoid possible circular reasoning, we try to develop it from the ground. The following is only a sketch.

Step 1. Let $\mathcal{V}$ be the familiy of Lebesgue integrable functions that satisfies your proposition. Prove that $\mathcal{V}$ is a $||\cdot||_{1}$-closed vector subspace of $L^{1}$.

Step 2. Prove that $1_{(a,b)}\in\mathcal{V}$ for any $-\infty<a<b<\infty$.

Step 3. Prove that $1_{A}\in\mathcal{V}$ for any Lebesgue measurable set $A$ with $\lambda(A)<\infty$. (It follows from outer-regularity: $\lambda(A)=\inf\sum_{n=1}^{\infty}\lambda(I_{n})$, where the infimum runs over all countable family of pairwisely disjoint open intervals $\{I_{n}\mid n\in\mathbb{N}\}$ that satisfies $A\subseteq\cup_{n}I_{n}$. In this step, note that $\sum_{n=N+1}^{\infty}\lambda(I_{n})$ is small when $N$ is large. More explicitly, choose such covering $\{I_{n}\mid n\in N\}$ for $A$ such that $\sum_{n}\lambda(I_{n})$ approximates $\lambda(A)$ well. Then further approximate $1_{A}\approx\sum_{n=1}^{N}1_{I_{n}}$ in $||\cdot||_{1}$-norm.)

Step 4. $f\in\mathcal{V}$ for any simple functions in $L^{1}$.

Step 5. $\mathcal{V}=L^1$

Danny Pak-Keung Chan
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