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I don't want too much of a hint, so won't display my particular example.

Given an unbounded function $\phi$. I want to compute its Lebesgue integral over $(a,b)$. It blows up to infinity at $b$ but is defined elsewhere. Being unbounded it is not Riemann integrable. The integral converges. The function is monotone increasing. I want a sequence of bounded functions that converge to $\phi$. Would $\phi_n$ the restriction of $\phi$ to $(a,b-\frac{1}{n})$ be a good choice, or does the domain of the functions in the sequence have to be $(a,b)$ if I want to use standard convergence theorems (such as dominated, monotone convergence etc)?

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    yes. my book doesn't define riemann integrability for unbounded functions. thus im avoiding the fact that if the improper integral converges it will be the same as the lebesgue.2012-11-29

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You could try and use $\phi_n = \begin{cases}\phi & (a,b-1/n)\\ 0 & [b-1/n,b) \end{cases} = \phi \cdot \chi_{(a,b-1/n)}$

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    @cap You may be interested in this example http://math.stackexchange.com/questions/222485/compute-lim-n-to-infty-int-0n-left1-fracx2n-rightne-x-dx/222490#222490 though here $b$ is $\infty$ and hence the corresponding sequence is f_n(x) = \left(1 + \dfrac{x}{2n} \right)^n e^{-x} \chi_{[0,n]} = \begin{cases} \left(1 + \dfrac{x}{2n} \right)^n e^{-x} & x \in [0,n]\\ 0 & x > n\end{cases}2012-11-29