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Let $f:[a,b]\to \mathbb{R}$ be a piecewise function such that $$f(x) = \left\{ \begin{array}{c} g(x) \hspace{1cm} a\le x<\alpha \\ h(x) \hspace{1cm} \alpha \le x \le b \end{array} \right. $$ where $\alpha \in (a,b)$. If we need to find the definite integral $$\begin{align} \int_{a}^{b} f(x) dx\end{align}$$ Is it equal to $$\int_a^{\alpha}g(x)dx + \int_{\alpha}^b h(x)dx$$

or since $g(x)$ in not defined for $\alpha$, do we need to use limit as the upper bound of the first integral approaches $\alpha$ i.e. $$\lim_{h\to 0^+}\int_a^{\alpha - h}g(x)dx$$
Here is an example where the integral is computed without using any limit.

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    @John in the link I have attached in the question, the function integrated is discontinuous in the interval in which it is being integrated. – 2017-02-15
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    @John I think you misinterpreted my question. I haven't defined it as you said. I have edited the question to make it more clear. – 2017-02-15
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    @John The Riemann integral does *not* need the function to be continuous: even functions with countably many discontinuities at Riemann-integrable. – 2017-02-15
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    @John: Continuity is not essential for Riemann integrability. More technically a bounded function is Riemann integrable if and only if its discontinuities form a set of Lebesgue measure zero. For your function, the right approach is to split the integral as a sum of two integrals over corresponding intervals as you have done. No need of limits near $\alpha$. – 2017-02-15
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    You are right. Sorry. It's been many years since I studied Lesbegue integral. I deleted my comments – 2017-02-15
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    @ParamanandSingh But why don't we use limit near $\alpha$? Since the interval is open near $\alpha$ we cant include it in first interval. – 2017-02-16
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    You mean to say that $\int_a^{\alpha}f(x)\,dx$ does not make sense? Why do you think so? – 2017-02-16
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    Note that your function $f$ is defined and bounded on entire interval $[a, b]$ and hence we can talk about the Riemann integral $\int_{a}^{b}f(x)\,dx$. Whether the integral exists or not is altogether a different matter. But given your specific function we can show that $\int_{a}^{\alpha}f(x)\,dx$ and $\int_{\alpha}^{b}f(x)\,dx$ exist and therefore $\int_{a}^{b}f(x)\,dx$ exists and $$\int_{a}^{b}f(x)\,dx = \int_{a}^{\alpha}f(x)\,dx + \int_{\alpha}^{b}f(x)\,dx$$ – 2017-02-16
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    @ParamanandSingh It does make sense as the integrand is $f(x)$ but in my case its $g(x)$. Since $g(x)$ is not defined for $\alpha$, it doesn't make sense (at least to me) to evaluate the integral as $G(\alpha)-G(a)$ where $G$ is any antiderivative of $g$. Maybe my intuition is wrong so you can correct me. – 2017-02-16
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    Note that we are not calculating the integral of $g$ here but rather the integral of $f$ here. The reason $g$ comes into picture is because of the way your $f$ is defined. – 2017-02-16
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    In case you are wondering here is the simple theorem which forms the basis of solution to all such questions. Suppose $f, g$ are bounded on $[a, b]$ and assume that $f(x) = g(x)$ except for a *finite* number of values of $x$ in $[a, b]$. Then either both the integrals $\int_{a}^{b}f(x)\,dx$ and $\int_{a}^{b}g(x)\,dx$ exist and have same value or both of them don't exist. – 2017-02-16
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    @ParamanandSingh So for this particular case we can use $g$. So we can have upper bound of an integral as a value for which the integrand in not defined? Because in this case $\alpha$ is evaluated in the second interval i.e. using function $h$. If that is the case then your theorem looks fine to me for evaluating the integral. – 2017-02-16

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From the comments I understood that your main concern is that $g(\alpha)$ is not defined and hence we can't talk of the symbol $$\int_{a}^{\alpha}g(x)\,dx$$ You are right here. But we have to talk of $\int_{a}^{\alpha}f(x)\,dx$ and not about integral of $g$ here.

The problem is handled by introducing another function $a(x)$ on $[a,\alpha]$ such that $a(x) = g(x)$ for $x \in [a, \alpha)$ and $a(\alpha) = h(\alpha)$. Now we can talk of the integral $\int_{a}^{\alpha}a(x)\,dx$ and note that we have $f(x) = a(x)$ if $x\in [a, \alpha]$ and $f(x) = h(x)$ if $x \in [\alpha, b]$ and then we can write $$\int_{a}^{b}f(x)\,dx = \int_{a}^{\alpha}a(x)\,dx + \int_{\alpha}^{b}h(x)\,dx$$ provided that both integrals on the right exist.

In reality we are not given functions like $g(x), h(x)$ but rather expressions in $x$ like $x + 1, 2x^{2} + \sin x$ in place of $g(x), h(x)$ and for these functions we do know that they are defined for the intermediate point $\alpha$.

Still suppose that there is a function $f$ which is defined and bounded on $[a, b)$ and we don't have any definition of $f$ for point $b$. Then as per definition of Riemann integral we can't talk about $\int_{a}^{b}f(x)\,dx$. What can we do now? Well we can define $f(b)$ to take any value (as per our wish) and then then talk of $\int_{a}^{b}f(x)\,dx$. As I mentioned in my comments, the values of a function at a finite number of points in an interval do not affect the existence/value of the Riemann integral of the function, hence we are at a liberty to choose any value of $f(b)$ and talk about integral of $f$ on $[a, b]$. This is precisely the way we think about integrals like $$\int_{0}^{1}\cos (1/t)\,dt$$ Limits are used for a different setup which goes by the name of Improper Riemann Integral. This setup is used when either the interval of integration is unbounded or if the function is unbounded on a bounded interval. Examples of both types are $$\int_{0}^{\infty}\frac{\sin x}{x} \,dx = \lim_{t \to \infty}\int_{0}^{t}\frac{\sin x}{x}\,dx,\, \int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}} = \lim_{t \to 1^{-}}\int_{0}^{t}\frac{dx}{\sqrt{1 - x^{2}}}$$

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    I understand now. One last question: is my approach using limits incorrect? – 2017-02-16
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    @Parth: Using limits is an extra burden, but it will work. – 2017-02-16
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    @Parth: I added some more details about the usage of limits for evaluating integrals. – 2017-02-16