Introduction

Suppose you play a game. We model the outcome of the game with the random variable $X$ where the probability of winning or losing is $p_X(1)=p_X(-1)=1/2$. So it is a fair game and $\mathbb{E}(X)=0$.
Suppose now that we play multiple times. We denote by $X_i$ the outcome of the $i$th game. Suppose we bet a piece of gold each time we play. So after the first round we have $X_1$ goldpieces. After the second one we would have $X_1+X_2$.
More generally, after the $t$-th round we would have $\begin{align*} Y_t=\sum_{i=1}^t X_i \end{align*}$ pieces of gold.
The expected value of $Y_i$ in the $(i-1)$th round, i.e., after we know $Y_1$, …, $Y_{i-1}$ would be $Y_{i-1}$. Since if we lose, we would have $Y_{i-1}-1$ and if we would win, we would have $Y_{i-1}+1$.

In this example, the family of random variables $Y_i$ is called a simple random walk. This motivates the following more general definition.

Definition

A family of random variables $Y_0$, $Y_1$, … is called a discrete martingale if $\mathbb{E}(Y_i)<\infty$ for all $i$ and for the conditional expected value it holds that $\mathbb{E}(Y_{n+1}|Y_0,\dots,Y_n)=Y_n$.

Note that the first assumption basically says that the expected value does exist.

Towards continuity

So what is a continuous martingale? Let us try to generalize our random walk. Let $Y_i$ be as before a discrete random walk. Now define $B_{j/N}=\frac{1}{\sqrt{N}} Y_j$.
We need a scaling factor, such that it works. This has to do with bounding the variance and i do not want to explain it. Note that if we let $N\rightarrow\infty$ then this stuff converges to something which i will denote by $B_t$. It is also called Brownian motion, therefore the symbol $B$. Please also note that $t\mapsto B_t$ is almost surely a continuous function. Furthermore because of the scaling we have that $B_\lambda-B_\mu \sim\mathcal{N}(0,\lambda-\mu)$ is normally distributed with mean $0$ and variance $\lambda-\mu$, the differences between any two increments of Brownian motion are independent and $B_0=0$.

Definition

Let $Y_i$, $i\geq 0$ be a family of random variables, $\mathcal{F}_t=\{Y_i,\,0\leq i\leq t\}$ a filtration. Then $Y_i$, $i\geq 0$ is called a martingale if $\mathbb{E}(Y_i)<\infty$ for every $i\leq 0$ and furthermore $\mathbb{E}(Y_i|\mathcal{F}_t)=Y_t$ for every $i>t$.

Integrals

Let us integrate this Brownian motion stuff. I mean, integrate as in Riemann integration. You know, we just take a discrete integral of the random walk and then do this limit stuff as explained above and then get an integral of Brownian motion.
Since Brownian motion is a stochastic process, the integral will also be a random variable. Therefore we can characterize it by computing its distribution. Obviously $I=\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{i=1}^N B_{i/N}$ should be the integral $I=\int_0^1 B_t dt$, if it exists and if the last part of that notation even makes sense. We have to hack around a bit, since only Brownian increments are independent.
$\begin{align*} \sum_{i=1}^N Y_i=NX_1+(N-1)X_2+...+X_N=\sum_{i=1}^N (N-i+1)X_i\sim\mathcal{N}\left(0,\frac{1}{N}\sum_{i=1}^N (N-i+1)^2\right) \end{align*}$
Where the last part follows, since the $X_i$ are independent. Moreover $\frac{1}{N}\sum_{i=1}^N (N-i+1)^2=\frac{(N+1)(2N+1)}{6}$ as one can show by induction or a search in your favorite collection of mystic formulae.
$\begin{align*} I_n=\frac{1}{N}\sum_{i=1}^N B_{i/N}\sim\mathcal{N}\left(0,\frac{(N+1)(2N+1)}{6N^2}\right)=\mathcal{N}\left(0,\frac{1}{3}+\frac{1}{2N}+\frac{1}{6N^2}\right) \end{align*}$
Letting $N\rightarrow\infty$ we get $\int_0^1 B_t dt=I\sim\mathcal{N}(0,\frac{1}{3})$.
We just integrated Brownian motion, bitches.