Interchanging limits under Integral for monotone decreasing functions I want to proof a Lemma about exchanging limits under the integral sign and i'm having struggles to complete it. To begin with:
Let ${ g }_{ n }$ and g be monotone decreasing (${ g }:\quad \left[ a,b \right] \rightarrow \Re $), if $\lim _{ n\rightarrow \infty  }{ { g }_{ n } } =g$ for all $x\epsilon \left[ a,b \right] $ 
than $\lim _{ n\rightarrow \infty  }{ \int _{ a }^{ b }{ { g }_{ n }(x) }  } =\int _{ a }^{ b }{ g(x)dx } $
I know that this fact is true if ${ g }_{ n }\rightarrow g$ converges uniformly. So i somehow want to deduce the uniform convergence for monotone functions which converge pointwise so that i can use the proof that i already know.I know that for ${ g }_{ n }\rightarrow g$ point wise, $\lim _{  }{ (sup\left| { g }_{ n }-g \right| )=0 } $ implies uniform convergence. So i want to show $\lim _{  }{ (sup\left| { g }_{ n }-g \right| )=0 } $ for monotone decreasing functions and i failed to do so.
Any suggestions,tipps?
IS it even possible this way or do i have to use facts like Riemann integrable or something diffrent?
Thanks for any Help
 A: This question is quite old, nevertheless people might stumble across it using google (as I did) and get confused - therefore I would like to add to the above answer.
To phrase the reason short: The equation
$$ \lim_{n\to\infty}\int_a^b g_n(x) dx = \int_a^b \lim_{n\to \infty}g_n(x) dx $$
always holds if $(g_n)$ is a pointwise convergent sequence of monotonically decreasing functions on a fixed interval $[a,b]$, even if the convergence is not uniform. Therefore the above approach of first proving uniform convergence to then conclude (**) can be misleading. In a way, uniform convergence ist for most functions a too strong assumption for the interchangeability of Riemann integrals and limits.
I will give some more details now. Goal is to prove the following lemma:
Let $g_n:[a,b]\to\mathbb R$ be monotonic decreasing for all $n\in \mathbb N$. Further assume the series $(g_n)$ converges pointwise to a function $g:[a,b]\to\mathbb R$.
Then $g$ is monotonic decreasing (and therefore Riemann-integrable) and the equation
$$ \int_a^b g(x) d x  = \lim_{n\to\infty}\int_a^b g_n(x) dx $$
holds.
Proof:
The proof of monotonicity is straightforward and is left to the reader. To prove the integral equation we use, that we can explicitly control upper and lower Riemann sums and their difference for monotonic functions.
For a partition $Z$ of $[a,b]$ and a Riemann-integrable function $f:[a,b]\to\mathbb R$ we denote the upper and lower Riemann sums as $O_Z(f)$ and $U_Z(f)$. For monotonically decreasing functions $f$ it is then easy to check (telescope priniciple), that
$$ O_Z(f)-U_Z(f)\le |Z|(f(a)-f(b))(b-a) $$
holds.
Now let $\varepsilon>0$ and fix a partition $Z_\varepsilon=(x_0,\ldots,x_N)$ of $[a,b]$, such that
$$ |Z_\varepsilon|(g(a)-g(b))(b-a)<\frac\varepsilon3 \quad \mbox{and}\quad |Z_\varepsilon|(g_n(a)-g_n(b))(b-a)<\frac\varepsilon3 $$
hold for all $n\in\mathbb N$. This is possible, due to the convergence $g_n(a)\to g(a)$ and $g_n(b)\to g(b)$.
This leads to
\begin{align}
 \left|\int_a^b g(x)dx-\int_a^b g_n(x)dx\right| & \le \left|O_{Z_\varepsilon}(g) - \int_a^b g(x)dx \right| + \left|O_{Z_\varepsilon}(g_n) - \int_a^b g_n(x)dx \right| + |O_{Z_{\varepsilon}}(g)-O_{Z_{\varepsilon}}(g_n)| \\ &
\le |O_{Z_\varepsilon}(g)-U_{Z_{\varepsilon}}(g)|+|O_{Z_{\varepsilon}}(g_n)-U_{Z_\varepsilon}(g_n)| + |O_{Z_{\varepsilon}}(g)-O_{Z_{\varepsilon}}(g_n)| \\&< \frac 23\varepsilon + \sum_{k=0}^{N-1}|g(x_k)-g_n(x_k)|(x_{k+1}-x_k),\end{align}
where we use the explicit form of upper sums for monotonically decreasing functions in the last step. We can now choose $n_\varepsilon$, such that
$$ |g(x_k)-g_n(x_k)|<\frac{\varepsilon}{3N(x_{k+1}-x_k)} $$
for all $k=0,\ldots,N-1$ and $n\ge n_\varepsilon$. This finally yields
$$ \left|\int_a^b g(x)dx-\int_a^b g_n(x)dx\right| < \varepsilon \qquad \mbox{for alle}\ n\ge n_\varepsilon,$$
which proves convergence of the Riemann integrals.
A: Check the hypotheses of the lemma.  
If we just assume pointwise convergence but add the condition that the limit function $g$ is continuous, we can show that the convergence must be uniform and switching the limit and integral is permissible.
Since $g$ is uniformly continuous on $[a,b]$, for any $\epsilon > 0$  there is a $\delta > 0$ such that $|x-y| < \delta$ implies $|g(x) - g(y)| < \epsilon/2$.  Taking a partition $(a = x_0,x_1, \ldots, x_{n-1}, x_n = b)$ where each subinterval has length less than $\delta$,  we have 
$$\tag{*}-\frac{\epsilon}{2} < g(x_{j-1}) - g(x_{j}) < \frac{\epsilon}{2}$$ 
Any $x \in [a,b]$ belongs to some subinterval $[x_{j-1},x_j]$. By pointwise convergence, there exists $N \in \mathbb{N}$, depending on $x_1,\ldots,x_{n-1}$ and $\epsilon$ but not $x$ such that for $n > N$,
$$ g(x_j) -\frac{\epsilon}{2}< g_n(x_j) , \quad g_n(x_{j-1}) < g(x_{j-1}) + \frac{\epsilon}{2}, $$
and, since $g_n$ and $g$ are monotone decreasing and $x_{j-1} \leqslant x \leqslant x_j$, it follows that
$$\begin{align}g(x_{j}) - g(x_{j-1}) - \epsilon/2 &< g_n(x_{j}) - g(x_{j-1}) \\&\leqslant g_n(x) - g(x) \\&\leqslant  g_n(x_{j-1}) - g(x_{j}) \\&< g(x_{j-1}) - g(x_{j}) + \epsilon/2\end{align}$$
Applying  (*) to the LHS and RHS we get for all $x \in [a,b]$ and $n > N$,
$$-\epsilon < g_n(x) - g(x) < \epsilon$$
Thus, the convergence $g_n \to g$ is uniform on $[a,b]$ and
$$\tag{**}\lim_{n \to \infty} \int_a^bg_n(x) \, dx = \int_a^bg(x) \, dx$$
Without assuming continuity of $\mathbf{g}$
Since $g$ is monotone it is discontinuous for at most countably many points where one-sided limits exist.  If there are only a finite number of discontinuities then the above proof can be applied on every interval where $g$ is continuous and the result (**) holds.
If $g$ has infinitely many discontinuity points, then it is possible that (**) is not true.  I have not yet found a counterexample or proof to the contrary.
