Sets of Measure Zero.

Heard the saying - What happens in Las Vegas, stays in Las Vegas? Sets of measure zero are kind of like that.

For example, say $f(x),g(x)$ are functions that are equal to each other in measurable space $E$, except on a subset $N$. Say a given measure on ``disagreeable'' space $N$ is equal to zero. Now $N$ is like Las Vegas and we are given license to forget about what happens in this space $N$ and assert that $f(x)=g(x)$ a.e where a.e stands for almost everwhere. To be more precise (just to prevent extra point from being leaked out of your exam paper!), we need to state $f(x)=g(x)$ a.e[$\mu$]. That is we need to specify which measure.

Mathematically, \begin{equation} \mu\{x:f(x)\neq g(x)\} = 0 \end{equation}

where $x \in N$. The above criteria specifies the points of $N$. Here $f$ ~ $g$ and it is not too difficult to prove that this is an equivalence relation.

Reflexive property: Clearly $f$ ~ $f$. The disagreeable set $N$ in this case is a null set and measure of null set is $0$. $f=f$ on the whole set. Symmetric: Clearly $f$ ~ $g$ also means that set of disagreeable space remains same when we switch $f$ to the right. Reflexive: Say $f$ ~ $g$ and let $N_1$ be the disagreeable set. Say $g$ ~ $h$ and let $N_2$ be the disagreeable set, then $N_1 \cup N_2$ where $f,g,h$ are not equal to each other. Hence $f$ ~ $h$.

Note all the above statements were made based on a property that $f=g$. Generally, we don't have to be specific about a property. Above assertions and proofs hold in a more abstract sense, that is for any property $P$.

R&C - Lebesgue's Dominated Convergence Theorem

Dominated Conv Theorem

If \(f \in \mathcal{L}(\mu)\), then \[\begin{equation} \left|\int_X f d\mu \right| \leq \int_x|f| d\mu \end{equation}\]

Proof uses the previously proven identity -

If \(f\) is a complex measurable function on \(X\), there is a complex measurable function \(\alpha\) on \(X\) such that \(|\alpha|=1\) and \(f=\alpha|f|\). This is an extension of property of complex numbers.

Start off by setting \(z=\int_X f d\mu\). Then there is another complex number \(\alpha\) such that \(|\alpha|=1\) and \(\alpha z= |z|\).

Let \(u\) be real part of \(\alpha f\). Then, \(u \leq |\alpha f|=|f|\). Hence,

\[\begin{align*} \left| \int_X f d\mu\right| = |z|=\alpha z=\alpha \int_X f d\mu \\ = \int_X \alpha f d\mu = \int_X u d\mu \leq \int_X |f| d\mu \end{align*}\] Suppose \(\{f_n\}\) is a sequence of complex measurable functions on \(X\) such that \[\begin{equation} f(x) = lim_{n \rightarrow \infty} f_n(x) \end{equation}\] exists for every \(x \in X\). If there is a function \(g \in \mathcal{L^1(\mu)}\) such that \[\begin{equation} |f_n(x)| \leq g(x) \text{ } (n=1,2,\cdots|x \in ) \end{equation}\] then \(f \in \mathcal{L^1(\mu)}\), \[\begin{equation} lim_{n \rightarrow \infty}\int_X |f_n - f| d\mu = 0 \end{equation}\] and \[\begin{equation} lim_{n \rightarrow \infty} \int_X f_n d\mu = \int_X f d\mu \end{equation}\]

Clear that \(|f| \leq g\). Since \(f_n\) are measurable, the limit \(f\) is measurable, \(f \in \mathcal{L^1(\mu)}\). \[\begin{align*} |f-f_n| \leq |f|+|f_n| \leq 2g \end{align*}\] This means, \(2g - |f_n - f|\) is a sequence of functions whose range is in \([0,\infty]\). Hence, precondition to satisfy Fatou’s lemma is satisfied. This yield \[\begin{align*} \int_X 2g d\mu \leq lim_{n \rightarrow \infty}inf \int_X(2g-|f_n-f|)d\mu \\ =\int_X 2g d\mu + lim_{n \rightarrow \infty}inf\left( - \int_X |f_n - f| d\mu\right) \\ = \int_X 2g d\mu - lim sup_{n \rightarrow \infty} \int_X |f_n - f| d\mu \end{align*}\] Taking advantage of finiteness of \(\int 2g d\mu\), \[\begin{align*} lim sup_{n \rightarrow \infty}\int_X |f_n - f| d\mu \leq 0 \end{align*}\] If sequence of nonnegative real numbers fails to converge to \(0\), then its upper limit is positive. Then above equation implies \[\begin{align*} lim _{n \rightarrow \infty}\int_X |f_n - f| d\mu = 0. \end{align*}\] Hence, \[\begin{align*} lim_{n \rightarrow \infty} \int_X f_n d\mu = \int_X f d\mu \end{align*}\]