Proof for integrability of uniformly converging function sequence Assume that, for each $n, f_{n}$ is an integrable function on $[a, b] .$ If $\left(f_{n}\right) \rightarrow f$ uniformly on $[a, b]$, prove that $f$ is also integrable on this set.
My attempt: We know that for  all $n$ and for each $\epsilon > 0$ there exists some partition $P_{\epsilon}$ s.t. $\underbrace{U(f_n, P_{\epsilon})}_{\text{upper sum}}-\underbrace{L(f_n, P_{\epsilon})}_{\text{lower sum}} < \epsilon$ (this partition can  be different for each $f_n$).
Let $M_k$ denote the supremum of the kth subinterval and $m_k$ the respective infimum of the according $f_n$ and let $M, m$  belong to $f$. Then, for a given $\epsilon > 0$ there exists some $N > 0$ s.t. for all $n \geq  N$  we find that $(\forall  x)\space|f(x_n)-f(x)| < \epsilon$ by the uniform convergence. Now let's consider such a $f_n$ with the corresponding $P_\epsilon:$
$\displaystyle\sum_{k = 0}^{n}(M_k-\epsilon)\cdot l <\displaystyle\sum_{k = 0}^{n}M\cdot l<\displaystyle\sum_{k = 0}^{n}(M_k+\epsilon)\cdot l \Longleftrightarrow U(f_n, P_\epsilon)-\epsilon \cdot l\cdot n < U(f,P_\epsilon) < U(f_n, P_\epsilon)+\epsilon \cdot l\cdot n  $
$\displaystyle\sum_{k = 0}^{n}(m_k-\epsilon)\cdot l <\displaystyle\sum_{k = 0}^{n}m\cdot l<\displaystyle\sum_{k = 0}^{n}(m_k+\epsilon)\cdot l \Longleftrightarrow L(f_n, P_\epsilon)-\epsilon \cdot l\cdot n < L(f, P_\epsilon) < L(f_n, P_\epsilon)+\epsilon \cdot l\cdot n  $
where I used the uniform convergence for the respective supremum/infimum in each subinterval. However, when I  subtract the lower from the upper inequality, I get nonsense ($\epsilon < \ldots < \epsilon$) so I was wondering where my mistake lies.
 A: Your mistake lies in dealing with the inequalities: When subtracting the lower from the upper, the inequalitise change, i.e. you get:
$ L(f_n, P_\epsilon)-\epsilon_1 \cdot l\cdot n < L(f, P_\epsilon) < L(f_n, P_\epsilon)+\epsilon_1 \cdot l\cdot n$
$ \Longleftrightarrow  -L(f_n, P_\epsilon)-\epsilon_1 \cdot l\cdot n <- L(f, P_\epsilon) < -L(f_n, P_\epsilon)+\epsilon_1 \cdot l\cdot n \space\space (1)$
Notice that the epsilon for the partition $P_\epsilon$ and $\epsilon_1$ can be different. Also, your $M$ and $m$ for $f$ might be a bit confusing since this notation conveys the notion that it's the same for every subinterval, what is indeed not the case. Now you can add $(1)$ to obtain:
$ U(f,P_\epsilon)-L(f, P_\epsilon) < U(f_n, P_\epsilon)+\epsilon_1 \cdot l\cdot n -L(f_n, P_\epsilon)+\epsilon_1 \cdot l\cdot n \space\space  = U(f_n, P_\epsilon)-L(f_n, P_\epsilon)  +  2\epsilon_1 \cdot l\cdot n $
$< \epsilon + 2\epsilon_1 l n$. Now, given any $\xi$, you could choose $\epsilon = \dfrac{\xi}{2}$ and $\epsilon_1 = \dfrac{\xi}{4\cdot l\cdot n}.$
A: Basically, you have already done the main work, you just have to substract the right terms: You have shown $U(f, P_{\varepsilon})<U(f_n, P_{\varepsilon})+\varepsilon$ and $L(f_n, P_{\varepsilon})-\varepsilon <L(f, P_{\varepsilon})$. This shows
$U(f, P_{\varepsilon})-L(f, P_{\varepsilon})<U(f_n, P_{\varepsilon})+\varepsilon-(L(f_n, P_{\varepsilon})-\varepsilon)=U(f_n, P_{\varepsilon})-L(f_n, P_{\varepsilon})+2\varepsilon<3\varepsilon$,
as $U(f_n, P_{\varepsilon})-L(f_n, P_{\varepsilon})<\varepsilon$ by assumption. As $\varepsilon >0$ was chosen arbitrarily, this shows that $f$ is integrable.
Generally, if you want a good upper bound for a difference $A-B$, then you need an upper bound for $A$ and a lower bound for $B$; e.g., if you know that $A_1<A<A_2$ and $B_1<B<B_2$, then you get $A-B<A_2-B_1$ (but you do not want to substract the whole inequalities).
