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Suppose $f\colon \mathbb{R} \to \mathbb{R}$ is (Lebesgue) integrable and $f=0$ outside of $[-1, 1]$. Define $f_{n}(x)=f(x+\frac{1}{n})$. Does $f_{n} \to f$ in measure? Does $f_{n} \to f$ in $L^{1}(\mathbb{R})$?

I know that convergence in $L^{1}(\mathbb{R})$ implies convergence in measure. I also can prove that in the case $f$ were to be continuous then we have convergence in $L^{1}(\mathbb{R})$ as follows:

For any $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $$|x-y|< \frac{1}{N} \quad \implies \quad |f(x)-f(y)|< \frac{\varepsilon}{3}.$$

Now for $n \ge N$, $$|x - (x + \frac{1}{n})|= \frac{1}{n}< \frac{1}{N} \quad \implies \quad |f(x)-f(x+\frac{1}{n})|<\frac{\varepsilon}{3}.$$

We then have \begin{align*} \|f - f_{n}\|_{1} &=\int_{\mathbb{R}} |f(x)-f(x+\frac{1}{n})| \\ &=\int_{-1-\frac{1}{n}}^{1}|f(x)-f(x+\frac{1}{n})| \quad \text{ since }f(x)=0 \text{ outside } [-1, 1]\\ &< \frac{\varepsilon}{3} \int_{-1-\frac{1}{n}}^{1} \\ &=\frac{\varepsilon}{3}(1 - (-1-\frac{1}{n}))\\ &=\frac{\varepsilon}{3}(2+\frac{1}{n})\\ &\le \varepsilon. \end{align*} Hence $f_{n} \to f$ in $L^{1}(\mathbb{R})$.

How do I prove the result for $f$ not continuous?

John Smith
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1 Answers1

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What you have used is not continuity but uniform continuity of $f$.

Let $\epsilon >0$. There exists a continuous function $g$ with compact support such that $\|f-g\|<\epsilon$. A simple change of varoiables shows tht $\|f_n-g_n\|<\epsilon$ where $g_n(x)=x+\frac 1 n$. $g$ is uniformly continuous so $g_n \to g$ in $L^{1}$. To finish the proof note that $$\|f_n-f\|\leq \|f_n-g_n\|+\|g_n-g\|+\|g-f\|.$$

geetha290krm
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