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If $(f_n)$ is a sequence of functions in $L^1$ that converge in measure to a function $f$, and if $(f_n)$ is uniformly integrable, then $f$ is also in $L^1$ and $\lim_{n\to\infty} \int f_n d\mu = \int f d\mu$.

If $(f_n)$ is a sequence of measurable functions that converge pointwise to a function $f$ and there exists an integrable function $g$ such that $|f_n| \leq g$ for all $n$, then $\lim_{n\to\infty} \int f_n d\mu = \int f d\mu$.

If $f$ is a measurable function on a set $E$ with finite measure, then for every $\epsilon > 0$, there exists a continuous function $g$ on $\mathbb{R}^n$ and a subset $E_\epsilon \subset E$ where $\mu(E \setminus E_\epsilon) < \epsilon$ such that $f = g$ on $E_\epsilon$.

Given a sequence of non-negative measurable functions $(f_n)$, then $\int \liminf_{n\to\infty} f_n d\mu \leq \liminf_{n\to\infty} \int f_n d\mu$.

Given a sequence of measurable functions $(f_n)$ that converge almost everywhere to a function $f$ on a set $E$ with finite measure, for every $\epsilon > 0$, there exists a subset $E_\epsilon \subset E$ with measure $\mu(E \setminus E_\epsilon) < \epsilon$ where the convergence of $(f_n)$ to $f$ is uniform on $E_\epsilon$.

If $(f_n)$ is a sequence of non-negative measurable functions increasing pointwise to a function $f$, then $\lim_{n\to\infty} \int f_n d\mu = \int f d\mu$.

For a locally integrable function $f$ on $\mathbb{R}^n$, at almost every point $x$, the limit of the average value of $f$ over spheres (or cubes) centered at $x$ with radius (or side) shrinking to zero is equal to $f(x)$.