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Let $a_n < c < b_n$ and $a_n, b_n → c$ as $n → ∞ $.

Suppose $ f : \mathbb{R} → \mathbb{R} $ is continuous.

Show that (Possibly using similar proof to FTC...?) $$ \lim_{n \to\infty }\frac{1}{{b_n - a_n}}\int_{{\,a_n}}^{{\,b_n}}{{f\left(x\right)\,dx}} = f(c).$$

jimmy_thomas33
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3 Answers3

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Note that $$\min_{a_n\leq x\leq b_n}f(x)\leq\frac1{b_n-a_n}\int_{a_n}^{b_n}f(x)\ \mbox{d}x\leq\max_{a_n\leq x\leq b_n}f(x).$$ Show that the left and right converge to $f(c)$ using the fact that $f$ is continuous and then use the squeeze theorem.

SmileyCraft
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  • How would I go about showing the left hand side tends to $ f(c) $ as $n → ∞$, not sure I'd know how to do this from continuity – jimmy_thomas33 Oct 24 '19 at 22:53
  • Apply the definition of continuity directly. For $\varepsilon>0$ there exists $\delta$ such that for all $c-\delta<x<c+\delta$ we have $f(c)-\varepsilon<f(x)<f(c)+\varepsilon$. Then there exists $N$ such that for all $n>N$ we have $c-\delta<a_n,b_n<c+\delta$. – SmileyCraft Oct 24 '19 at 22:58
  • I understand that, you've shown the left and right side converge to $f(c)$. What does the "Then there exists $N$ such that for all $n>N$ we have $c−δ<a_n,b_n<c+δ$" Then prove sorry? Thanks – jimmy_thomas33 Oct 24 '19 at 23:30
  • Well then for $n>N$ we have for all $a_n<x<b_n$ that $c-\delta<x<c+\delta$ and therefore $f(c)-\varepsilon<f(x)<f(c)+\varepsilon$ and therefore $f(c)-\varepsilon<\min_{a_n<x<b_n}f(x)\leq\max_{a_n<x<b_n}f(x)<f(c)+\varepsilon$. – SmileyCraft Oct 25 '19 at 12:07
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If you already acknowledge Integral Mean Value, then $\dfrac{1}{b_{n}-a_{n}}\displaystyle\int_{a_{n}}^{b_{n}}f(x)dx=f(\eta_{n})$ for some $\eta_{n}$ in between $a_{n}$ and $b_{n}$. Taking $n\rightarrow\infty$ and squeezing $a_{n},b_{n}\rightarrow c$ to get $\eta_{n}\rightarrow c$ and continuity of $f$ implies $f(\eta_{n})\rightarrow f(c)$.

Edit:

By @SmileyCraft answer, since $f$ is continuous, the maxima and minima are attained: \begin{align*} f(\alpha_{n})\leq\int_{a_{n}}^{b_{n}}f(x)dx\leq f(\beta_{n}) \end{align*} for some $a_{n}\leq\alpha_{n},\beta_{n}\leq b_{n}$. Now use Intermediate Value Theorem to get some $a_{n}\leq\eta_{n}\leq b_{n}$ such that \begin{align*} \dfrac{1}{b_{n}-a_{n}}\int_{a_{n}}^{b_{n}}f(x)dx=f(\eta_{n}). \end{align*}

user284331
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  • In the question, it's assumed I do not know about the mean value theorem for integrals. If I proved this theorem, then did what you said above, would that suffice or would a better way be proving similar to the FTC proof? Thanks – jimmy_thomas33 Oct 24 '19 at 23:00
  • @jimmy_thomas33, look at the edited post. – user284331 Oct 24 '19 at 23:18
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Let $\varepsilon > 0$ be given. Since $f$ is continuous at $c$, there exists $\delta > 0$ so that for all $x$ with $|x-c| < \delta$ we have $|f(x)-f(c)| < \varepsilon$.

Now also we have $a_n \to c$ and $b_n \to c$, so there exists $N \in \mathbb N$ so that for all $n \ge N$ we gave $c-a_n<\delta$ and $b_n - c< \delta$.

Fix $n \ge N$. Then for all $x \in [a_n,b_n]$ we have $|x-c| < \delta$. [Indeed, if $x<c$ then $a_n \le x \le c$ so $|x-c| < \delta$; and similarly if $x>c$. Finally, it is also true when $x=c$.] Thus, for all $x \in [a_n,b_n]$ we have $|f(x)-f(c)| < \varepsilon$. Compute $$ \frac{1}{b_n-a_n}\int_{a_n}^{b_n} f(x)\;dx - f(c) = \frac{1}{b_n-a_n}\int_{a_n}^{b_n} \big(f(x)-f(c)\big)\;dx < \frac{1}{b_n-a_n}\int_{a_n}^{b_n} \varepsilon\;dx = \varepsilon $$ and similarly $$ \frac{1}{b_n-a_n}\int_{a_n}^{b_n} f(x)\;dx - f(c) > -\varepsilon $$

Finally, then: for any $\varepsilon > 0$ there is $N \in \mathbb N$ so that if $n > N$ we have $$ \left|\frac{1}{b_n-a_n}\int_{a_n}^{b_n} f(x)\;dx - f(c)\right| < \varepsilon . $$

Note: we do not require continuity of $f$ everywhere, merely at the one point $c$.

GEdgar
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