Let f: [a,b] -> R. Given f is integrable and bounded below, show 1/f is integrable.
To show that 1/f is integrable, we need to prove that it satisfies the integrability condition.
Since f is integrable, we know that it satisfies the Riemann integrability condition. This means that for any given epsilon > 0, there exists a partition P such that the upper Riemann sum minus the lower Riemann sum for f over that partition is less than epsilon.
Now, let's consider the function 1/f. We want to show that it also satisfies the Riemann integrability condition.
First, note that since f is bounded below, there exists a constant M > 0 such that f(x) >= M for all x in [a, b].
Now, let epsilon > 0 be given. We need to find a partition P' of [a, b] such that the upper Riemann sum minus the lower Riemann sum for 1/f over that partition is less than epsilon.
Let's construct the partition P' by dividing [a, b] uniformly into n subintervals. We choose n large enough so that each subinterval has length less than epsilon/M.
Now, let's calculate the upper Riemann sum and the lower Riemann sum for 1/f over the partition P'.
The upper Riemann sum is defined as the sum of the supremum of 1/f on each subinterval multiplied by the length of that subinterval.
The lower Riemann sum is defined as the sum of the infimum of 1/f on each subinterval multiplied by the length of that subinterval.
Since f is bounded below by M, we have 1/f <= 1/M for all x in [a, b]. Therefore, the supremum of 1/f on each subinterval is 1/M.
Also, since f is bounded below by M, we have 1/M <= 1/f for all x in [a, b]. Therefore, the infimum of 1/f on each subinterval is 1/f(x), where x is the point that achieves the infimum.
Now, let's calculate the difference between the upper Riemann sum and the lower Riemann sum for 1/f over the partition P'.
The difference is given by:
(1/M) * (b - a) - (1/f(x)) * (b - a)
Notice that (1/M) * (b - a) is a constant, and (1/f(x)) * (b - a) is the sum of the reciprocals of f(x) over all subintervals.
Since f is integrable, we know that the difference between the upper Riemann sum and the lower Riemann sum for f over P is less than epsilon.
Therefore, we can choose a partition P of [a, b] for f such that the difference is less than epsilon.
Thus, we have shown that for the chosen partition P of f, the difference between the upper Riemann sum and the lower Riemann sum for 1/f over P' is less than epsilon.
Hence, 1/f is integrable.