From sum to integral

Question

I have to show that $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=-n}^nf(j/n)=\int_{-1}^1f(t)\mathrm{d}t,$$ where $f:[-1,1]\to\mathbb{R}$ is a continuous and positive function. Intuitively the result makes sense, but I'd like to prove it formally. Any help is much appreciated!

Answer 1

Approach 1

I'll rewrite it on the form you seem to want it based on your comments. Notice that

$$\frac{1}{n}\sum_{j=-n}^nf\left(\frac{j}{n}\right)=\frac{1}{n}\sum_{j=0}^{2n}f\left(\frac{j-n}{n}\right)=\frac{1}{n}\sum_{j=0}^{2n}f\left(-1+\frac{j}{n}\right).$$

In particular, if we set $m=2n$, then

$$\frac{1}{n}\sum_{j=-n}^nf\left(\frac{j}{n}\right)=\frac{2}{m}\sum_{j=0}^mf\left(-1+\frac{2j}{m}\right).$$

Now let $n\to\infty$ (consequently making $m\to\infty$) to get your result.

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Approach 2

Another way, which I would say is preferable, is simply to notice immediately that $\mathcal{P}_n=\left\{\frac{j}{n}:j\in\{-n,\dots,n\}\right\}$ forms a partition of $[-1,1]$, and so, as $\lvert\mathcal{P}_n\rvert\to0$ as $n\to\infty$, and the function is Riemann integrable, the result follows from the definition.

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Approach 3 (Overkill)

This third approach I'm including just for the fun of it, and is not how you should be doing this. Anyways, it is a way you can avoid working directly with the definition. Indeed notice first that

$$\int_{-1}^1f(t)~\mathrm{d}t=\sum_{j=-n}^{n-1}\int_{\frac{j}{n}}^{\frac{j+1}{n}}f(t)~\mathrm{d}t$$

and that

$$\frac{1}{n}f\left(\frac{j}{n}\right)=\int_{\frac{j}{n}}^{\frac{j+1}{n}}f\left(\frac{j}{n}\right)~\mathrm{d}t.$$

From this we get that

$$\left\lvert\int_{-1}^1f(t)~\mathrm{d}t-\frac{1}{n}\sum_{j=-n}^{n-1}f\left(\frac{j}{n}\right)\right\rvert=\left\lvert\sum_{j=-n}^{n-1}\int_{\frac{j}{n}}^{\frac{j+1}{n}}\left(f(t)-f\left(\frac{j}{n}\right)\right)~\mathrm{d}t\right\rvert\leq\sum_{j=-n}^{n-1}\int_{\frac{j}{n}}^{\frac{j+1}{n}}\left\lvert f(t)-f\left(\frac{j}{n}\right)\right\rvert~\mathrm{d}t.$$

Now let $\varepsilon>0$ and choose $N\in\mathbb{Z}^+$ so that

$$\lvert f(x)-f(y)\rvert<\varepsilon$$

whenever

$$\lvert x-y\rvert<\frac{1}{N}$$

(this can be done as $f$ is uniformly continuous on $[-1,1]$). It follows that for all $n\geq N$,

$$\left\lvert\int_{-1}^1f(t)~\mathrm{d}t-\frac{1}{n}\sum_{j=-n}^{n-1}f\left(\frac{j}{n}\right)\right\rvert\leq\sum_{j=-n}^{n-1}\int_{\frac{j}{n}}^{\frac{j+1}{n}}\left\lvert f(t)-f\left(\frac{j}{n}\right)\right\rvert~\mathrm{d}t\leq\sum_{j=-n}^{n-1}\int_{\frac{j}{n}}^{\frac{j+1}{n}}\varepsilon~\mathrm{d}t=\varepsilon.$$

Thus

$$\lim_{n\to\infty}\frac{1}{n}\sum_{j=-n}^{n-1}f\left(\frac{j}{n}\right)=\int_{-1}^1f(t)~\mathrm{d}t,$$

and consequently

$$\lim_{n\to\infty}\frac{1}{n}\sum_{j=-n}^{n}f\left(\frac{j}{n}\right)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{j=-n}^{n-1}f\left(\frac{j}{n}\right)+\frac{f(1)}{n}\right)=\int_{-1}^1f(t)~\mathrm{d}t+0=\int_{-1}^1f(t)~\mathrm{d}t.$$