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I was solving this problem , to check the measurability of this piecewise continuous function, $$f(x) = \begin{cases} 2, \ \ 0 < x < 1 \\ \pi, \ \ 1 < x < 2 \\ 1.5, \ \ 2 < x < 3 \end{cases}$$

i thought of using the lemma that if $O$ is any open set , then if $f^{-1}(O)$ is measurable then $f$ is measurable but i can't proceed by this way.

Next , i tried sketching the graph of this function and i got it to be asimple function , as it takes finite no. of function values and as simple functions are measurable , $f$ is also measurable.

Is this correct , any other kind of method?

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    Also i think it will be Lebesgue integrable, is the integral value = $2 + \pi + 1.5 $ , is it so? – 2017-02-21

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Note that we can write $$f = g_1\mathbf{1}_{I_1} + g_2\mathbf{1}_{I_2} + g_3\mathbf{1}_{I_3},$$

where each $g_i$ is a constant function, hence measurable, and each $I_i$ is an open interval, hence measurable. Now use the fact that indicator functions of measurable sets are measurable and the fact that products and sums of measurable functions are measurable.

And you are correct about the lebesgue integral because $\lambda(I_i) = 1$, hence (identifying constant functions with their values) $$\lambda(f) = \sum_i g_i\lambda(I_i) = 2 + \pi + 1.5.$$

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    Here $\textbf{1}$ represents the Characteristic function? – 2017-02-22
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    I think then only the above works as each $I_{i}$ are open intervals of measure $1$ , measurable and hence Characteristic function is measurable . – 2017-02-22
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    Yes, it's the characteristic, or indicator, function. – 2017-02-22