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 $\varepsilon &gt;0$ 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?
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
$$
f_{1,1}(x) = \begin{cases}1 &amp; x \in [0,1/2)\\ 0 &amp; x \in [1/2, 1] \end{cases},~~
f_{1,2}(x) = \begin{cases}0 &amp; x \in [0,1/2)\\ 1 &amp; x \in [1/2, 1] \end{cases}
$$
Generally, for $k &gt; 1, 1 \leq l &lt; 2^k$, we define
$$
f_{k,l}(x) = \begin{cases}1 &amp; x \in [(l-1)2^{-k},l 2^{-k})\\ 0 &amp; \text{else} \end{cases}, ~~ f_{k,2^k}(x) = \begin{cases}1 &amp; x \in [1-2^{-k},1]\\ 0 &amp; \text{else} \end{cases},
$$
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?
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.