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Let $f$ be Riemann integrable on $[a,b]$ then $$ \int_a^b f(x) \, \mathrm{d}x = \lim\limits_{n \to ∞} \dfrac{b-a}{n} \sum\limits_{r=1}^n f \left(a+\dfrac{r(b-a)}{n} \right) $$

I want to prove this, $$ \lim\limits_{n \to ∞} \dfrac{1}{n} \sum\limits_{t=an}^{bn} f \left(\dfrac{t}{n}\right) = \int_a^b f(x) \, \mathrm{d}x$$

Here's my attempt: Let $a,b \in \mathbb{Z}$ and $K=b-a$ then $$ I= \int_a^b f(x) \, \mathrm{d}x = \lim\limits_{n \to ∞} \dfrac{K}{n} \sum\limits_{r=1}^n f \left(a+\dfrac{rK}{n} \right) = \lim\limits_{n \to ∞} \dfrac{K}{n} \sum\limits_{r=1}^n f \left(\dfrac{an+rK}{n}\right) $$

Since $1≤r≤n$ so if $K>0$ then $an + K ≤ an+rK=t ≤ bn$

$$1) \quad I= \lim\limits_{n \to ∞} \dfrac{K}{n} \sum\limits_{t=an+K}^{bn} f \left(\dfrac{t}{n}\right) =\lim\limits_{n \to ∞} \dfrac{K}{n}\left(\sum\limits_{t=an}^{bn}f\left(\dfrac{t}{n}\right)- \sum\limits_{t=an}^{an+K-1} f\left(\dfrac{t}{n}\right) \right)= \lim\limits_{n \to ∞}\dfrac{K}{n}\sum\limits_{t=an}^{bn}f\left(\dfrac{t}{n}\right) $$

Similarly, $1≤r≤n$ so if $K<0$ then $an + K ≥ an+rK=t ≥ bn$ so,

$$2) \quad I = \lim\limits_{n \to ∞}\dfrac{K}{n}\sum\limits_{t=bn}^{an}f\left(\dfrac{t}{n}\right)$$

Almost there, but I don't know how to get rid of $K$ outside the sum in $1$ and $2$. If $K$ is gone, my result is proved. Any help?

William
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  • I'm a bit confused on the notation with the bounds of your summation. Since $a,b$ don't have to be integers, how can one use $an$ and $bn$ as the bounds of a summation? – Alan Jun 16 '21 at 15:48
  • @Alan read just below the result where I say "Let... " :) – William Jun 16 '21 at 15:55
  • How did you go from $\sum\limits_{r=1}^n f \left(\frac{an+rK}{n}\right)$ to $\sum\limits_{t=an}^{bn}f\left(\frac{t}{n}\right)$? If $f(x)=1$ then the first sum is $,n,$ and the second sum is $,bn-an=(b-a)n$. – Somos Jun 16 '21 at 18:08
  • @Somos You forget that $n \to ∞$. Of course they're not the same if $n$ is a finite number. The first sum translates to a different sum with $r=an + K$ to $r=bn$ (see above), your second sum (that starts from $r=an$ to $r=bn$) is obtained from this "different sum" by adding and subtracting the missing terms between $an$ and $an+K$ and using the fact that $\lim\limits_{n \to ∞}$ – William Jun 16 '21 at 22:29
  • @Somos The spirit of the idea is same that in $$ \sum\limits_{r=a}^n f\left(\dfrac{t}{n}\right) = \int_0^1 f(x) , \mathrm{d}x$$ That is— You do not need to start from $r=1$, you can start from any $a \in \mathbb{Z}$ – William Jun 16 '21 at 22:32
  • Again, suppose that $f(x)$ is a constant function, then your step from $\sum_{r=1}^n$ to $\sum_{t=an}^{bn}$ differs by the factor $K=b-a$ which explains why your factor of $K$ disappears. – Somos Jun 16 '21 at 23:41
  • @Somos I see your point. That is a problem. Although I can't where could I have went wrong. All I did was a substitution to force it into $t/n$ but apparently that substitution is actually what's causing the problem. Would you be kind enough to help me find that mistake? – William Jun 17 '21 at 05:32
  • @Somos Also, if not substitution then how should I prove the result? I don't see any other way :'( – William Jun 17 '21 at 05:42

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