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Let Mn and mn denote the max and min of g on [n, n+1]. Note that g does not attain its max or min at the endpoints unless its constant on the interval, in which case we're done. Also, Mn is strictly increasing and mn is strictly decreasing. This is clear since we are assuming g is not constant, in which case on the interval [xn, xn + 1], where g(xn) = Mn with xn in (n, n+1), there is an x for which g(x) > Mn. By assumption, this x > n+1, so Mn+1 > Mn. Similarly for mn+1 < mn.

Hence the sequences of maxima and minima have limits, say M and m. Say M - m = c > 0. For any (epsilon) e > 0, there exists N such that M - Mn < e and mn - m < e for all n > N. Now let xn attain Mn and yn attain mn. WLOG yn > xn (opposite works the same). Because we have

'[; \int{x_n}^{x_n + 1} g(t) dt = g(x_n) ;]', and g(t) < M for all t here, we also have '[; |\int{yn}^{x_n + 1} g(t) - g(x_n) dt| < \epsilon ;]', whence '[; |\int{yn}^{x_n + 1} g(t) - g(y_n) dt| > c - 3\epsilon ;]' by triangle inequality. But similarly since g(t) > m, we have '[; |\int{x_n+1}^{y_n+1} g(t) - g(y_n) dt| < \epsilon ;]'. Adding up the last two inequalities gives the contradiction.