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?
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 $0$.