Show that $\lim\limits_{n\to \infty}\sum\limits_{k=-n^2}^{n^2}\left|\int_{k/n}^{(k+1)/n}f(x)\,dx\right| = \int_{-\infty}^\infty |f(x)|\, dx$

Question

Let $f\in L^1(\mathbb{R})$. Show that $$\lim_{n\to \infty}\sum_{k=-n^2}^{n^2}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right| = \int_{-\infty}^\infty |f(x)|\ dx.$$

Attempt at Solution: I figure approximating $f$ in the $L^1$ norm by a continuous function with compact support will be helpful, so let's take $f\in C_0(\mathbb{R})$. If $f$ doesn't change sign over an interval $I$, then $|\int_I f\ dx| = \int_I|f|\ dx$. If $f$ is continuous on a compact set, I want to say that $f$ changes sign only finitely often there. Basically, we only pick up errors on the LHS when we integrate over a region where $f$ changes sign. Our regions get smaller as $n\to \infty$ and the continuity of $f$ will ensure that our errors get smaller in the limit. I'm just not sure how to make this rigorous.

Answer 1

I would make this a comment if I had enough rep. I'll be happy to delete it if you want. For continuous compactly supported $f$, use uniform continuity to ensure that for $n\geq N_\epsilon$, the errors (i.e. sign changes of $f$) occur only in intervals where $|f|$ is no greater than $\epsilon$. Then each error term is bounded by $\epsilon\cdot 1/n$ and you have $O(n)$ of them (since we're working in a bounded interval). So the error terms contribute only $\epsilon$ to the LHS and $\epsilon$ was arbitrary.

Answer 2

It is true for continuous fuctions in $ L^1$. To see why it holds for $ L^1$, you can always approximate an $ L^1$ $ f $ with a continuous function $ g $ such that $ |f-g|_{L^1} \le \epsilon $, $\epsilon > 0$

Answer 3

One sided is clear: \begin{align*} \sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right|&\leq\sum_{k=-n^{2}}^{n^{2}}\int_{k/n}^{(k+1)/n}|f(x)|dx\\ &=\int_{-n}^{n+1/n}|f(x)|dx, \end{align*} so \begin{align*} \limsup_{n\rightarrow\infty}\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right|\leq\int_{-\infty}^{\infty}|f(x)|dx. \end{align*} Now turn the other one. Given $\epsilon>0$, find a $\varphi\in C_{0}^{\infty}({\bf{R}})$ such that $\|f-\varphi\|_{L^{1}({\bf{R}})}<\epsilon$. Let $M\in{\bf{N}}$ be such that $\text{supp}(\varphi)\subseteq\{|x|\leq M\}$, then \begin{align*} \int_{-\infty}^{\infty}|f(x)|dx-\epsilon&<\int_{-\infty}^{\infty}|\varphi(x)|dx=\int_{-M}^{M}|\varphi(x)|dx, \end{align*} and for all $n\geq M$, then \begin{align*} \sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}\varphi(x)dx\right|&=\sum_{k=-M}^{M}\sum_{j=0}^{n-1}\left|\int_{k+j/n}^{k+(j+1)/n}\varphi(x)dx\right|. \end{align*} But \begin{align*} \sum_{j=0}^{n-1}\left|\int_{k+j/n}^{k+(j+1)/n}\varphi(x)dx\right|&=\sum_{j=0}^{n-1}\left|\int_{k+j/n}^{k+(j+1)/n}\eta'(x)dx\right|,~~~~\varphi:=\eta'\\ &=\sum_{j=0}^{n-1}|\eta(k+(j+1)/n)-\eta(k+j/n)|\\ &\rightarrow\int_{k}^{k+1}|\eta'(x)|dx,~~~~\text{rough reasoning by bounded variation}\\ &=\int_{k}^{k+1}|\varphi(x)|dx. \end{align*} Keeping in mind that $k$ varies only for finitely many, so for large $n$, \begin{align*} \int_{-M}^{M}|\varphi(x)|dx-\epsilon&<\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}\varphi(x)dx\right|\\ &\leq\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}(\varphi(x)-f(x))dx\right|+\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right|\\ &\leq\|f-\varphi\|_{L^{1}({\bf{R}})}+\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right|, \end{align*} so \begin{align*} \int_{-\infty}^{\infty}|f(x)|dx-3\epsilon<\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right|, \end{align*} this shows that \begin{align*} \int_{-\infty}^{\infty}|f(x)|dx\leq\liminf_{n\rightarrow\infty}\sum_{k=-n^{2}}^{n^{2}}\left|\int_{k/n}^{(k+1)/n}f(x)dx\right|. \end{align*}