Vtyjkyuk

This is the proof for the Squeeze theorem for integrals.

But in the last line, it is written that "Now, we can easily see that...", but I just cannot find out how $$|S(f,P_n)-S(f,P_n-1)|<frac1n$$ from the previous two inequalities.

I think the argument below should work for the proof, even if it doesn't actually produce the stated bound. However, let me know if you have any issues -- perhaps I've misunderstood something about what the proof is trying to do.

Observe that

beginalign* leftvert S(f,dotmathcal P_n+1) - S(f,dotmathcal P_n) rightvert &le leftvert S(omega_n+1,dotmathcal P_n+1) - S(alpha_n,dotmathcal P_n) rightvert \
&le leftvert S(omega_n+1,dotmathcal P_n+1) - Z_n+1 rightvert + leftvert A_n - S(alpha_n,dotmathcal P_n) rightvert + leftvert Z_n+1 - A_nrightvert \
& le leftvert S(omega_n+1,dotmathcal P_n+1) - Z_n+1 rightvert + leftvert A_n - S(alpha_n,dotmathcal P_n) rightvert + int_a^b left( omega_n+1 - alpha_n right).
endalign*

This should give us enough to deduce that $S(f, dotmathcalP_n)$ is Cauchy, albeit not the specific bound given.

The first inequality follows from the fact that $alpha_n le f le omega_n+1$ (but see parenthetical remark at the end), the second inequality is the triangle inequality, and the third from the fact that $vertint fvert le intvert f vert$.

One potentially problematic part here is that the last term in the last expression above doesn't exactly correspond to the condition we assumed, but this should be rectified by appropriate relabelling of $omega_varepsilon$ and $alpha_varepsilon$. (This also guarantees the first inequality, which would not work if $S(alpha_n,dotmathcal P_n) > S(omega_n+1,dotmathcal P_n+1)$.) That is, the conditions stated guarantee that we can choose a pair of sequences $alpha_n$ and $omega_n$ such that $alpha_n le f le omega_n+1$ and $int_a^b (omega_n+1 - alpha_n) < 1/n$.