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Let $f(x)$ an $g(x)$ be integrable functions over $[a,b]$ and let $\alpha$ be a point of $[a,b]$ if $f(x) = g(x)$ for all $x\neq \alpha$, then $$\int_{[a,b]}f(x)dx=\int_{[a,b]}g(x)dx.$$

So is it okay to say, since $f(x)=g(x)\Rightarrow f(x)-g(x)=0$. Then also given that both functions are integrable I can say that for $f(x)$ to be Riemann integral there is a $|T-K|<\varepsilon$ if $\max(x_i-x_{i-1})<\delta$. For $g(x)$ to be Riemann integral there is a $|T-K|<\varepsilon$ if $\max(x_i-x_{i-1})<\delta$ and then set them equal to each other?

I'm really confused where my next step is, and finding delta is becoming a little abstract. Please can someone help me solve or give me a nice push in the right direction?

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    $\begingroup$ Presumably you are using the Riemann definition of the integral. When you make a partition of $[a,b]$, consider the subinterval which contains $\alpha$ separately to the rest of the partition. $\endgroup$ – preferred_anon Jun 26 '15 at 0:36
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    $\begingroup$ Try breaking things into 3 integrals. The first from $a$ to $\alpha - \epsilon$. The second from $\alpha - \epsilon$ to $\alpha + \epsilon$. Then the third from $\alpha + \epsilon$ to $a$. Try to bound each of these integrals individually and then send $\epsilon$ to zero. $\endgroup$ – cnick Jun 26 '15 at 0:38
  • $\begingroup$ Thank you that helps a lot! $\endgroup$ – Whitts42 Jun 27 '15 at 4:51
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As cnick suggested: split both integrals as $$ \int_a^b = \int_a^{\alpha-\epsilon}+\int_{\alpha-\epsilon}^{\alpha+\epsilon} + \int_{\alpha+\epsilon}^b \tag{1} $$ The goal is to show that $$\int_{[a,b]}f(x)dx - \int_{[a,b]}g(x)dx = 0 \tag{2}$$ With the split as in $(1)$, the difference of integrals becomes $$\int_{\alpha-\epsilon}^{\alpha+\epsilon}f(x)dx - \int_{\alpha-\epsilon}^{\alpha+\epsilon}g(x)dx = \int_{\alpha-\epsilon}^{\alpha+\epsilon}(f(x)-g(x))dx \tag{3}$$ This can be estimated: for example, let $M=\max(\sup|f-g|)$; then the right hand side of $(3)$ is at most $2M\epsilon$. Thus, $$\left|\int_{[a,b]}f(x)dx - \int_{[a,b]}g(x)dx\right|\le 2M\epsilon$$ and since $\epsilon$ can be arbitrarily small, the difference is actually $0$.

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