If a function f:[ a,b] to R has finitely many points of discontinuity in [ a,b] , then f is integrable on [ a,b].
True or false with full explanation
Solution:
True.
Proof:
Suppose "c \\in(a, b)" . Let "\\varepsilon>0" . Since f is bounded, there exists M>0 such that "|f(x)| \\leq M" for all "x \\in[a, b]" . Let "\\delta=\\varepsilon \/ 12 M" . Since f is continuous on"[a, c-\\delta]" and "[c+\\delta, b]" , f is Riemann-integrable on "[a, c-\\delta]" and "[c+\\delta, b]" , so there exists partitions "P_{1}=\\left\\{a=x_{0}<\\cdots<x_{n}=c-\\delta\\right\\} of [a, c-\\delta]" and "P_{2}=\\left\\{c+\\delta=y_{0}<\\cdots<y_{m}=c-\\delta\\right\\}" of "[c+\\delta, b]" such that
"U\\left(P_{1}, f\\right)-L\\left(P_{1}, f\\right)<\\frac{\\varepsilon}{3} \\quad \\text { and } \\quad U\\left(P_{2}, f\\right)-L\\left(P_{2}, f\\right)<\\frac{\\varepsilon}{3}"
Consider the partition of [a, b] given by "P=P_{1} \\cup P_{2}" . Then
"\\begin{aligned}\n\nU(P, f) &=\\sum_{i=1}^{n} M_{i} \\Delta x_{i}+2 \\delta \\sup _{x \\in[c-\\delta, c+\\delta]} f(x)+\\sum_{j=1}^{m} M_{j} \\Delta y_{j} \\leq U\\left(P_{1}, f\\right)+2 M \\delta+U\\left(P_{2}, f\\right) \\\\\n\nL(P, f) &=\\sum_{i=1}^{n} m_{i} \\Delta x_{i}+2 \\delta \\inf _{x \\in[c-\\delta, c+\\delta]} f(x)+\\sum_{j=1}^{m} m_{j} \\Delta y_{j} \\geq L\\left(P_{1}, f\\right)-2 M \\delta+L\\left(P_{2}, f\\right)\n\n\\end{aligned}"
and so
"\\begin{aligned}\n\nU(P, f)-L(P, f) & \\leq\\left[U\\left(P_{1}, f\\right)-L\\left(P_{1}, f\\right)\\right]+4 M \\delta+\\left[U\\left(P_{2}, f\\right)-L\\left(P_{2}, f\\right)\\right] \\\\\n\n&<\\frac{\\varepsilon}{3}+4 M \\times \\frac{\\varepsilon}{12 M}+\\frac{\\varepsilon}{3}=\\varepsilon\n\n\\end{aligned}"
Hence f is Riemann integrable.
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