Riemann Integrals converging to $0$ Let's assume that $\{f_n\}$ is a sequence of real,continuous,nonnegative functions on $[0,1]$ such that $$\int_0^1f_n(x)dx\overset{n\to\infty}{\longrightarrow} 0$$
I need to prove, or disprove that there exist points  $x_o\in[0,1]$ such that $\lim\limits_{n\to\infty}f_n(x_0)=0$.
I feel that there must exist such points. So, suppose on the contrary that $\forall x\in[0,1] ,f_n(x)\nrightarrow0$. Taking an $x$ we can find a subsequence $\{f_{k_n}\}$ of $\{f_n\}$ such that $f_{k_n}(x)\geq\varepsilon$ for some $\varepsilon >0$ and $\forall n\in \mathbb N$. But how can this work? I can find an area around $x$ in which $f_{k_n}$ is away from zero , but what about when $n$ gets larger? The problem is that $f_{k_n}(y)$ can still converge to zero for $y\neq x$ in that area, and I am not getting a contradiction since I am working on a subsequence.
Do you think there can be a counterexample?
 A: This is false. To see why not, try constructing a counterexample in the following way (the counterexample I give is not continuous, but you can make it continuous with a modification).
For notational ease, I'll write this as a kind of array of functions, which you can make into a linear sequence in a clear way. Write $f_{0,1}(x) = 1$ identically, and
$$
f_{1,1}(x) = \begin{cases}1 & x \in [0,1/2)\\ 0 & x \in [1/2, 1] \end{cases},~~
f_{1,2}(x) = \begin{cases}0 & x \in [0,1/2)\\ 1 & x \in [1/2, 1] \end{cases}
$$
Generally, for $k > 1, 1 \leq l < 2^k$, we define
$$
f_{k,l}(x) = \begin{cases}1 & x \in [(l-1)2^{-k},l 2^{-k})\\ 0 & \text{else} \end{cases}, ~~ f_{k,2^k}(x) = \begin{cases}1 & x \in [1-2^{-k},1]\\ 0 & \text{else} \end{cases},
$$
For $k$ fixed, varying $l$ has the effect of shifting around the 'bump' of width $2^{-k}$. Writing $g_1 = f_{0,1}, g_2 = f_{1,1}, g_3 = f_{1,2}$ and so on, we have now a sequence of functions $g_n$ for which
$$
\int_0^1 g_n dx \rightarrow 0
$$
Can you see why there is no point $x_0 \in [0,1]$ for which $g_n(x) \rightarrow 0$? Moreover, do you see how to turn this into a counterexample for when the $g_n$ are continuous?
A: I think that it is possible that no such $x_0$ exists. Consider the following sequence of functions.
Let $f_1(x)=1$.
Let $f_2(x)$ be a function that takes value $1$ if the fractional part of $x$ is in interval $[0,1/2]$, value $0$, if it is in the interval $[1/2+\epsilon,1-\epsilon]$
for some small $\epsilon$, and is piecewise linear in between. Note that this function is periodic with period $1$. We then define $f_3(x)=f_2(x-1/2)$.
Next we define functions $f_4,f_5,f_6,f_7$. We let $f_4(x)$ be equal to one in the interval $[0,1/4]$, zero in the interval $[1/4+\epsilon_2,1-\epsilon_2]$, and again extend periodically and piecewise linearly.
We then define $f_5(x)=f_4(x-1/4)$, $f_6(x)=f_4(x-1/2)$, $f_7(x)=f_4(x-3/4)$.
Continue similarly next defining $f_8,\ldots,f_{15}$.
The integral of $f_n$ is close to $(1/2)^{\log_2 n}$ so tends to zero. Yet for each $x_0$ we have $f_n(x_0)=1$ for infinitely many values of $n$.
A: because $f_n: [0,1] \to \mathbb{R}_{\geq0}$, $f_n \to f$ in L1  $\implies f_n \to 0$ in measure.  This however is not enough.  You can come up with $f_n$ s.t. $f_n \notin L_\infty $ over any subset of [0,1] and have it converge to 0 in $L_1$  You can have subsequences with this property though.
A: Counterexample:
$$
{\rm f}_{\rm n}\left(x\right) \equiv {1 \over \left(x + n\right)^{2}} > 0\,,
\qquad\qquad
\int_{0}^{1}{\rm f}_{n}\left(x\right)
=
\int_{0}^{1}{{\rm d}x \over \left(x + n\right)^{2}}
=
{1 \over n\left(n + 1\right)}
$$
